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Write the equation of the parabola that passes through the points (2,28(-2,28), (1,0(-1,0), and (5,0(5,0). Write your answer in the form y=a(xp)(xq)y = a(x - p)(x - q), where aa, pp, and qq are integers, decimals, or simplified fractions.

Full solution

Q. Write the equation of the parabola that passes through the points (2,28(-2,28), (1,0(-1,0), and (5,0(5,0). Write your answer in the form y=a(xp)(xq)y = a(x - p)(x - q), where aa, pp, and qq are integers, decimals, or simplified fractions.
  1. Identify X-Intercepts: We have the points (2,28(-2,28), (1,0(-1,0), and (5,0(5,0). The points (1,0(-1,0) and (5,0(5,0) are the x-intercepts of the parabola, so they give us the values of pp and qq directly.\newlinep=1p = -1 and q=5q = 5.
  2. Write Parabola Equation: Now we can write the equation of the parabola in the form y=a(xp)(xq)y = a(x - p)(x - q) using the values of pp and qq we found.\newliney=a(x(1))(x5)y = a(x - (-1))(x - 5)\newliney=a(x+1)(x5)y = a(x + 1)(x - 5)
  3. Find Value of a: Next, we need to find the value of aa. We will use the point (2,28)(-2,28) to do this. Substitute x=2x = -2 and y=28y = 28 into the equation y=a(x+1)(x5)y = a(x + 1)(x - 5).\newline28=a(2+1)(25)28 = a(-2 + 1)(-2 - 5)
  4. Simplify and Solve: Simplify the equation to solve for aa.28=a(1)(7)28 = a(-1)(-7)28=7a28 = 7aa=287a = \frac{28}{7}a=4a = 4
  5. Final Equation: Now that we have found a=4a = 4, we can write the final equation of the parabola.y=4(x+1)(x5)y = 4(x + 1)(x - 5)

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