What is a quadratic function with a maximum at #(3, 125)# and roots at #-2# and #8#?

Question

James Allen · Accepted Answer

#y=-5x^2+30x+80#

The two roots are #-2# and #8#. This means that the function can be written as

#y=a(x+2)(x-8)#

where #a# is some constant.

Plug in #(x,y)rarr(3,125)# to find what #a# must be.

#125=a(3+2)(3-8)#

#125=a(5)(-5)#

#125=-25a#

#a=-5#

Thus, the equation of the parabola is #y=-5(x+2)(x-8)#, or #y=-5x^2+30x+80#.

graph{-5, 12, -20, 140]} = 5x^2 + 30x+80

Greyson Brown · Answer

undefined A quadratic function in standard form is given by $ f(x) = ax^2 + bx + c $. Given that the function has a maximum at (3, 125) and roots at -2 and 8, we can use this information to find the specific quadratic function.

Since the vertex of the quadratic function is the maximum point, we know that the vertex form of the quadratic function is $ f(x) = a(x - h)^2 + k $, where $ (h, k) $ is the vertex.

Given that the maximum occurs at (3, 125), we have $ h = 3 $ and $ k = 125 $.

Also, since the roots of the quadratic function are -2 and 8, we know that the factors of the quadratic function are $ (x + 2) $ and $ (x - 8) $.

By expanding $ (x + 2)(x - 8) $, we get $ x^2 - 6x - 16 $.

Now, substituting the values of $ h $ and $ k $ into the vertex form, we have:

$$ f(x) = a(x - 3)^2 + 125 $$

Expanding this, we get:

$$ f(x) = a(x^2 - 6x + 9) + 125 $$

$$ f(x) = ax^2 - 6ax + 9a + 125 $$

Now, we know that $ a $ is determined by the coefficient of $ x^2 $, which is 1 in our expanded form.

Therefore, $ a = 1 $.

So, the quadratic function is:

$$ f(x) = x^2 - 6x + 9 + 125 $$

$$ f(x) = x^2 - 6x + 134 $$