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A parabola opening up or down has vertex $(0,0)$ and passes through $(-4,-4)$ . Write its equation in vertex form. $\newline$ Simplify any fractions.

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Write a quadratic function from its vertex and another point

Full solution

Q. A parabola opening up or down has vertex

(0,0)

and passes through

(-4,-4)

. Write its equation in vertex form.

\newline

Simplify any fractions.

Vertex Form: What is the vertex form of the parabola? $\newline$ Vertex form of a parabola: $y = a(x - h)^2 + k$
Equation at ( $0$ , $0$ ): What is the equation of a parabola with a vertex at $(0, 0)$ ? $\newline$ Substitute $0$ for $h$ and $0$ for $k$ in vertex form. $\newline$ $y = a(x - 0)^2 + 0$ $\newline$ $y = ax^2$
Substitute $(-4, -4)$ : $y = ax^2$ $\newline$ Replace the variables with $(-4, -4)$ in the equation. $\newline$ Substitute $-4$ for $x$ and $-4$ for $y$ . $\newline$ $-4 = a(-4)^2$ $\newline$ $-4 = 16a$
Solve for $a$ : $-4 = 16a$ $\newline$ Solve for $a$ . $\newline$ $-4 = 16a$ $\newline$ $-\frac{4}{16} = a$ $\newline$ $-\frac{1}{4} = a$
Equation with $a = -\frac{1}{4}$ : $y = ax^2$ $\newline$ What is the equation of the parabola if $a = -\frac{1}{4}$ ? $\newline$ Substitute $-\frac{1}{4}$ for $a$ . $\newline$ $y = (-\frac{1}{4})x^2$ $\newline$ Vertex form of the parabola: $y = -\left(\frac{1}{4}\right)x^2$

More problems from Write a quadratic function from its vertex and another point

Question

h

is defined over the real numbers. This table gives a few values of

h

\newline

\begin{tabular}{lllll}

\newline

x

-6

1

-6

01

-6

001

-5

9

\newline

h(x)

-0

25

-0

74

-0

98

-1

0

\newline

\newline

What is a reasonable estimate for

\lim _{x \rightarrow-6} h(x)

\newline

1

\newline

-6

\newline

-2

\newline

-1

\newline

(D) The limit doesn't exist

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