Question

How would I solve? Please help!

Answer 1

Answer:

• intercepts: (-3, 0), (5, 0)
• vertex: (1, -16)

Step-by-step explanation:

a. Compare the given equation to the standard form to see what the coefficients are:

  ax^2 +bx +c = 0

  x^2 -2x -15 = 0

Comparing these, we see that ...

  a = 1, b = -2, c = -15

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b. Using the quadratic formula we find the intercepts to be ...

  $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x=\dfrac{-(-2)\pm\sqrt{(-2)^2-4(1)(-15)}}{2(1)}\\\\=\dfrac{2\pm\sqrt{64}}{2}=1\pm4\\\\x=\{-3,5\}$

The x-intercepts of the function are (-3, 0) and (5, 0).

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c. The midpoint between the vertices is ...

  x = (-3+5)/2

  x = 1

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d. The value of the function for x=1 is ...

  y = 1^2 -2(1) -15

  y = -16

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e. The coordinates of the vertex are ...

  (x, y) = (1, -16)