Question

Please help! what is an equation of a parabola with x-intercepts at (2,0) and (-7,0) and which passes through the point (1,32)?​

Answer 1

Answer:

           f(x) = - 4x² - 20x + 56  

Step-by-step explanation:

f(x) = a(x - x₁)(x - x₂)     - factored form of the equation of the parabola with zeros x₁ and x₂

x-intercepts at (2,0) and (-7,0) means zeros: x₁=2 and x₂=-7

So:

f(x) = a(x - 2)(x + 7)     - factored form of the equation of the parabola with x-intercepts at (2,0) and (-7,0)

The parabola passing through point (1, 32) means if x=1 then f(x)=32

Then:

        32 = a(1 - 2)(1 + 7)  

        32 = a(-1)(8)  

        32 = - 8a

         a = - 4

Therefore the equation of a parabola with x-intercepts at (2,0) and (-7,0) and which passes through the point (1,32):

                           f(x) = -4(x - 2)(x + 7)  

Expanding to standard form:

     f(x) = -4(x - 2)(x + 7)

     f(x) = -4(x² + 7x - 2x - 14)  

     f(x) = -4x² - 20x + 56