Standard Form of a Quadratic Function
Introduction
A quadratic function is a polynomial of degree `2`. Any polynomial function in which the highest exponent of the variable(s) is two qualifies as a quadratic function. The word "Quadratic" comes from the word "Quad" which means square. When graphed, a quadratic function forms a U-shape graph called a parabola. We come across parabolas in various real-life scenarios like the path taken by a projectile, the path a rocket takes when it is launched, the arc of a bridge, etc.
What Is the Standard Form of a Quadratic Equation?
The standard form of a quadratic equation is `ax^2 + bx + c = 0`, where `a` is the leading coefficient and it is a non-zero real number.
The standard form of a quadratic function is derived from its parent function `f(x) = x^2`. Apart from the standard form of a quadratic equation, it can be written in other forms as stated below.
Various Forms of Quadratic Equations
Understanding the various forms of quadratic equations and how to convert between them is fundamental in algebra. Let's look into the standard form of a quadratic equation in detail and explore how to convert it into other forms.
Standard Form: `ax^2 + bx + c = 0`
In the standard form, `a`, `b`, and `c` are constants, and `a` must be non-zero. Here `c` represents the `y`-intercept of the parabola. This form is versatile and useful for solving quadratic equations using methods like factoring, completing the square, or using the quadratic formula.
Vertex Form: `a (x - h)^2 + k = 0`
The vertex form represents the quadratic equation in terms of its vertex, which is the turning point of the parabola. Here, `h` and `k` represent the `x`-coordinate and `y`-coordinate of the vertex, respectively. Converting from standard to vertex form typically involves completing the square.
Intercept Form: `a (x - p)(x - q) = 0`
The intercept form expresses the quadratic equation in terms of its `x`-intercepts (or roots). Here, `p` and `q` represent the `x`-values where the parabola intersects the `x`-axis. Converting from standard to intercept form usually involves factoring the quadratic expression.
Converting between these forms can involve algebraic techniques such as factoring, completing the square, or expanding expressions. Each form has its advantages depending on the context of the problem you're working on. Understanding how to convert between them allows for greater flexibility and insight into quadratic equations.
We can break down the standard form of a quadratic equation as follows:
\( ax^2 + bx + c = 0 \)
In this equation:
- ` a `, ` b `, and ` c ` are constants,
- ` x ` is the variable, and
- ` ax^2 ` represents the quadratic term,
- ` bx ` represents the linear term, and
- ` c ` is the constant term.
Please note that the direction of opening of a quadratic graph in standard form is determined by the sign of the coefficient ` a `. If ` a ` is positive, the parabola opens upwards, and if ` a ` is negative, the parabola opens downwards.
Standard Form of Quadratic Equation
Example `1`: Identify `a, b` and `c` in `2x^2 - 5x + 3 = 0`.
Solution:
In the given equation:
`a = 2`
`b = -5`
`c = 3`
Example `2`: Identify `a, b` and `c` in `-x^2 + 4x - 7 = 0`.
Solution:
In the given equation:
`a = -1`
`b = 4`
`c = -7`
Example `3`: Identify `a, b` and `c` in `3x^2 + 6x + 2 = 0`.
Solution:
In the given equation:
`a = 3`
`b = 6`
`c = 2`
These examples represent quadratic equations in standard form, where `a`, `b`, and `c` are constants, `x` is the variable, and the equation is set equal to zero.
Quadratic Form to Standard Form
To convert a quadratic into standard form `ax^2 + bx + c = 0` or how to write the quadratic function in standard form or to convert quadratic function to standard form, you'll typically need to perform algebraic operations to rearrange the equation. Depending on the initial form of the quadratic equation, the process may involve different steps.
Here's a general guideline on how to convert a quadratic to standard form:
Step `1`: Expand and simplify: If the quadratic equation is given in vertex form `(x - h)^2 + k` or intercept form `a(x - p)(x - q)`, expand and simplify the expression to obtain a quadratic expression in the form `ax^2 + bx + c`.
Step `2`: Combine like terms: Collect similar terms together to get the quadratic expression in the form `ax^2 + bx + c`.
Step `3`: Ensure the equation equals zero: The standard form of a quadratic equation has the form `ax^2 + bx + c = 0`, so ensure that the equation is set equal to zero.
Step `4`: Check the leading coefficient: Ensure that the coefficient `a` of the `x^2` term is not zero, as a quadratic equation with a zero coefficient for `x^2` would not be a quadratic equation.
Example: Convert the quadratic equation `2(x - 3)^2 - 8 = 0` into standard form.
Solution:
Step `1`: Expand and simplify:
Expand `(x - 3)^2` to obtain `x^2 - 6x + 9`.
Rewrite the equation as `2(x^2 - 6x + 9) - 8 = 0`.
Step `2`: Distribute and combine like terms:
Distribute the `2` into the expression to get `2x^2 - 12x + 18 - 8 = 0`.
Simplify to obtain `2x^2 - 12x + 10 = 0`.
Step `3`: Ensure the equation equals zero:
Make sure the equation is set equal to zero:
`2x^2 - 12x + 10 = 0`.
Now, the quadratic equation `2(x - 3)^2 - 8 = 0` is converted into standard form `2x^2 - 12x + 10 = 0`.
Quadratic Equation: Standard Form to Vertex Form
To convert a quadratic equation from standard form `(ax^2 + bx + c = 0)` to vertex form `(a(x - h)^2 + k = 0)`, we typically use the method of completing the square. Here's how to do it step by step:
Step `1`: Identify the coefficients:
In the standard form `ax^2 + bx + c = 0`, identify the values of `a`, `b`, and `c`.
Step `2`: Find the vertex:
The vertex of the parabola represented by the quadratic equation is given by the point `(h, k)`, where `h = -\frac{b}{2a}`. Calculate `h`.
Step `3`: Complete the square:
Rewrite the quadratic expression `ax^2 + bx` as `a(x^2 + \frac{b}{a}x)`.
Now, we'll add and subtract `(\frac{b}{2a})^2` inside the parentheses to complete the square.
This gives us:
`a\left(x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2\right)`
Step `4`: Factor and simplify:
Factor the quadratic expression inside the parentheses and simplify:
`a\left(\left(x + \frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2 \right)`
Step `5`: Write in vertex form:
Rearrange the expression to put it in vertex form:
`a\left(x + \frac{b}{2a}\right)^2 - a\left(\frac{b}{2a}\right)^2 + c = 0`
`a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c = 0`
Now you have the quadratic equation in vertex form `a(x - h)^2 + k = 0` where `h = -\frac{b}{2a}` and `k = -\frac{b^2}{4a} + c`.
Example: Convert the quadratic equation `x^2 + 4x - 3 = 0` from standard form to vertex form.
Solution:
Step `1`: Identify the coefficients:
In this equation, `a = 1`, `b = 4`, and `c = -3`.
Step `2`: Find the vertex:
The `x`-coordinate of the vertex, `h`, is given by `h = -\frac{b}{2a}`.
Substitute `a` and `b` into the formula:
`h = -\frac{4}{2(1)} = -2`
So, `h = -2`.
Step `3`: Complete the square:
Add and subtract `(\frac{4}{2})^2 = 4` inside the parentheses:
\(\left(x^2 + 4x + 4 - 4\right)\)
Step `4`: Factor and simplify:
Factor the quadratic expression inside the parentheses:
\(\left((x + 2)^2 - 4\right)\)
Step `5`: Write in vertex form:
Rearrange the expression to put it in vertex form:
\((x + 2)^2 - 4 -3 = 0\)
\((x + 2)^2 - 7 = 0\)
Now, the quadratic equation `x^2 + 4x - 3 = 0` is converted to vertex form `(x + 2)^2 - 7 = 0`, where the vertex is `(-2, -7)`.
Quadratic Equation: Vertex Form to Standard Form
To convert a quadratic equation from vertex form `a(x - h)^2 + k = 0` to standard form `ax^2 + bx + c = 0`, you can follow these general steps:
Step `1`: Expand the squared term: Expand `(x - h)^2` using the distributive property or the binomial theorem to obtain `x^2 - 2hx + h^2`.
Step `2`: Distribute the leading coefficient `a`: Multiply `a` to each term obtained from expanding the squared term.
Step `3`: Combine like terms: Simplify the expression by combining like terms.
Step `4`: Ensure the equation is in standard form: Ensure that the quadratic equation is in the standard form `ax^2 + bx + c = 0`, where `a`, `b`, and `c` are constants.
Example: Convert the quadratic equation `3(x - 2)^2 + 4 = 0` from vertex form to standard form.
Solution:
Step `1`: Expand the squared term: Use the distributive property or the binomial theorem to expand `(x - 2)^2`:
\( (x - 2)^2 = (x - 2)(x - 2) = x^2 - 4x + 4 \)
Step 2: Distribute the leading coefficient: Multiply each term inside the parentheses by the coefficient `a = 3`:
\( 3(x^2 - 4x + 4) + 4 \)
Step `3`: Simplify: Distribute `3` and combine like terms:
\( 3x^2 - 12x + 12 + 4 \)
\( 3x^2 - 12x + 16 \)
Step `4`: Ensure the equation is in standard form: Equate the expression from Step `3` to `0`.
\( 3x^2 - 12x + 16 = 0 \)
Now, the quadratic equation `3(x - 2)^2 + 4 = 0` is converted to standard form `3x^2 - 12x + 16 = 0`.
Quadratic Equation: Standard Form to Intercept Form
To convert a quadratic equation from standard form `ax^2 + bx + c = 0` to intercept form `a(x - p)(x - q) = 0`, you typically need to factor the quadratic expression and then rewrite it in intercept form. Here's how you can do it step by step:
Step `1`: Factor the quadratic expression: Factor the quadratic expression `ax^2 + bx + c` into the form `(x - p)(x - q)`, where `p` and `q` are the `x`-intercepts (roots) of the quadratic equation. You can use methods such as factoring, completing the square, or the quadratic formula to find the roots.
Step `2`: Write in intercept form: Once you have factored the quadratic expression, rewrite it in intercept form `a(x - p)(x - q) = 0`, where `p` and `q` are the `x`-intercepts.
Example: Convert the quadratic equation `x^2 - 5x + 6 = 0` from standard form to intercept form.
Solution:
Step `1`: Factor the quadratic expression:
We need to find two numbers that multiply to `6` and add to `-5`. Those numbers are `-2` and `-3`.
So, `x^2 - 5x + 6` factors into `(x - 2)(x - 3)`.
Step `2`: Write in intercept form:
Rewrite the factored expression in intercept form:
`(x - 2)(x - 3) = 0`.
Now, the quadratic equation `x^2 - 5x + 6 = 0` is converted to intercept form `(x - 2)(x - 3) = 0`, where the `x`-intercepts are `p = 2` and `q = 3`.
Quadratic Equation: Intercept Form to Standard Form
To convert a quadratic equation from intercept form `a(x - p)(x - q) = 0` to standard form `ax^2 + bx + c = 0`, you'll need to expand the factors and then simplify the expression. Here's how you can do it step by step:
Step `1`. Expand the factors: Use the distributive property or FOIL method to expand `(x - p)(x - q)` into the form `x^2 - (p + q)x + pq`.
Step `2`. Distribute the leading coefficient `a`: Multiply each term obtained from expanding the factors by the coefficient `a`.
Step `3`. Combine like terms: Simplify the expression by combining like terms.
Step `4`. Ensure the equation is in standard form: Ensure that the quadratic equation is in the standard form `ax^2 + bx + c = 0`, where `a`, `b`, and `c` are constants.
Example: Convert the quadratic equation `(x - 2)(x - 3) = 0` from intercept form to standard form.
Solution:
Step `1`: Expand the factors:
Use the FOIL method to expand `(x - 2)(x - 3)`:
\( (x - 2)(x - 3) = x^2 - 3x - 2x + 6 \)
\( = x^2 - 5x + 6 \)
Step `2`: Distribute the leading coefficient:
There's no leading coefficient `a` given in this case, so we'll assume `a = 1`.
Multiply each term by `a = 1`:
\( x^2 - 5x + 6 \)
Step `3`: Ensure the equation is in standard form:
The equation `x^2 - 5x + 6 = 0` is already in standard form.
Now, the quadratic equation `(x - 2)(x - 3) = 0` is converted to standard form `x^2 - 5x + 6 = 0`.
Real-Life Applications of Standard Form of Quadratic Equation
One real-life application of the standard form of a quadratic equation `ax^2 + bx + c = 0` is in physics, particularly in projectile motion.
When an object is thrown or projected into the air with a certain initial velocity, its height above the ground can be modeled by a quadratic equation due to the influence of gravity. The standard form of this quadratic equation is often used to analyze and predict the motion of the object.
Here's how the standard form of a quadratic equation applies to projectile motion:
`1`. Modeling the motion: The equation `ax^2 + bx + c = 0` can be used to represent the height `y` of the object at any given time `x`. The coefficient `a` represents the acceleration due to gravity, `b` represents the initial velocity of the object, and `c` represents the initial height or position of the object.
`2`. Predicting the maximum height: By analyzing the quadratic equation, we can determine the vertex of the parabola, which represents the maximum height reached by the object. This maximum height is an important parameter in projectile motion analysis.
`3`. Finding the time of flight: The quadratic equation can also be used to find the time taken by the object to reach the ground again. This is done by solving the equation for `x` when `y = 0`, which gives us the roots (or solutions) of the quadratic equation.
`4`. Optimizing trajectories: Engineers and designers use the principles of projectile motion to optimize the trajectories of projectiles such as rockets, missiles, and projectiles in sports like golf and baseball. By analyzing the standard form of the quadratic equation, they can optimize parameters such as launch angle and initial velocity to achieve desired outcomes.
Solved Examples
Example `1`: Convert the quadratic equation `3(x + 2)^2 - 5 = 0` into standard form.
Solution:
Expand `(x + 2)^2` to obtain `x^2 + 4x + 4`.
Rewrite the equation as:
`3(x^2 + 4x + 4) - 5 = 0`
Distributing the `3` into the expression to get:
`3x^2 + 12x + 12 - 5 = 0`
Simplify to obtain `3x^2 + 12x + 7 = 0`.
Make sure the equation is set equal to zero: `3x^2 + 12x + 7 = 0`
Now, the quadratic equation `3(x + 2)^2 - 5 = 0` is converted into standard form `3x^2 + 12x + 7 = 0`.
Example `2`: Convert the quadratic equation `x^2 + 6x + 9 = 0` from standard form to vertex form.
Solution:
In this equation, `a = 1`, `b = 6`, and `c = 9`.
Using the formula for the `x`-coordinate of the vertex: `h = -\frac{b}{2a}`.
Substitute `a` and `b` into the formula:
`h = -\frac{6}{2(1)} = -\frac{6}{2} = -3`
Rewriting the quadratic expression by completing the square:
\( (x^2 + 6x) + 9 \)
To complete the square, add and subtract `(\frac{b}{2a})^2`:
\( (x^2 + 6x + 9 - 9) + 9 \)
\( (x + 3)^2 - 9 + 9 \)
\( (x + 3)^2 \)
Now, the quadratic equation is in vertex form: `(x + 3)^2 = 0`, where `h = -3` and `k = 0`.
Example `3`: Given the vertex form equation: ` y = 4(x - 1)^2 - 3 `. Convert it into an equation in its standard form.
Solution:
Converting it to standard form:
\( y = 4(x^2 - 2x + 1) - 3 \)
\( y = 4x^2 - 8x + 4 - 3 \)
\( y = 4x^2 - 8x + 1 \)
The equation in standard form is `4x^2 - 8x + 1 = 0 `.
Example `4`: Given the quadratic equation in standard form: ` y = x^2 - 4x + 4 `. Convert it into an equation in its intercept form.
Solution:
We need to factor it into the intercept form.
\( x^2 - 4x + 4 = (x - 2)^2 \)
The only `x`-intercept in this case is ` x = 2 `.
So, the intercept form is: ` (x - 2)^2 = 0 ` or ` y = (x - 2)(x - 2) = 0 `.
Example `5`: Given the quadratic equation in intercept form: ` y = (x - 1)^2 `. Convert in into standard form.
Solution:
To convert it to standard form, we expand the expression:
\( y = x^2 - 2x + 1 \)
So, the equation in standard form is ` y = x^2 - 2x + 1 `.
Practice Problems
Q`1`. Convert the quadratic equation `4(x - 1)^2 + 9 = 0` into standard form.
- `4x^2 - 8x + 13 = 0`
- `4x^2 - 8x + 9 = 0`
- `4x^2 - 4x + 13 = 0`
- `4x^3 - 8x + 13 = 0`
Answer: a
Q`2`. Convert the quadratic equation `2x^2 + 4x - 3 = 0` from standard form to vertex form.
- `2(x + 1)^2 - 3 = 0`
- `2(x + 1)^2 - 5 = 0`
- `2(x + 1)^2 - 2 = 0`
- `2(x + 1)^2 + 4 = 0`
Answer: b
Q`3`. Given the vertex form equation: ` y = 2(x + 3)^2 - 6 `. Convert it into standard form.
- ` y = 2x^2 + 12x - 6 `
- ` y = 2x^2 - 12x + 12 `
- ` y = 2x^2 + 12x + 12 `
- ` y = 2x^2 + 12x + 3 `
Answer: c
Q`4`. Given the quadratic equation in standard form: ` y = -3x^2 + 6x - 2 `. Rewrite it in vertex form.
- ` y = -3(x + 1)^2 + 3 `
- ` y = -3(x - 1)^3 + 2 `
- ` y = 3(x - 1)^2 + 3 `
- ` y = -3(x - 1)^2 + 1 `
Answer: d
Q`5`. Given the quadratic equation in intercept form: ` y = (x - 2)(x - 5) `. Write it in standard form.
- ` y = x^2 - 7x + 10 `
- `y = x^2+7x+10`
- ` y = x^2 - 7x - 10 `
- ` y = x^2 - 7x + 5 `
Answer: a
Frequently Asked Questions
Q`1`. What is the standard form of a quadratic function?
Answer: The standard form of a quadratic function is given by ` y = ax^2 + bx + c `, where ` a `, ` b `, and ` c ` are constants, and ` a ` is not equal to zero.
Q`2`. Why is standard form important?
Answer: Standard form is important because it provides a clear and concise way to represent quadratic functions. It allows for easy identification of the coefficients ` a `, ` b `, and ` c `, which in turn can provide information about the shape, direction, and location of the quadratic graph.
Q`3`. How do you graph a quadratic function in standard form?
Answer: To graph a quadratic function in standard form, you can use techniques such as finding the vertex, axis of symmetry, and intercepts. Additionally, you can identify the direction of the graph (opening upwards or downwards) based on the sign of the coefficient ` a `.
Q`4`. Can a quadratic function in standard form have complex coefficients?
Answer: Yes, a quadratic function in standard form can have complex coefficients. For example, ` y = (3 + 2i)x^2 - (1 - i)x + (2 - 4i) ` is a quadratic function with complex coefficients.
Q`5`. How do you convert a quadratic function from standard form to vertex form?
Answer: To convert a quadratic function from standard form to vertex form, you can use the method of completing the square. This involves rewriting the quadratic expression as a perfect square trinomial plus or minus a constant.
Q`6`. What is the relationship between the standard form and the vertex form of a quadratic function?
Answer: The vertex form of a quadratic function `y = a(x - h)^2 + k` and the standard form `y = ax^2 + bx + c` are algebraically equivalent. The vertex form provides information about the vertex of the parabola, while the standard form offers information about the `y`-intercept, coefficients, and the concavity.
Q`7`. How do you determine the direction of the opening of a quadratic graph in standard form?
Answer: The direction of the opening of a quadratic graph in standard form is determined by the sign of the coefficient ` a `. If ` a ` is positive, the parabola opens upwards, and if ` a ` is negative, the parabola opens downwards.