x-intercepts of 2 and 7 and maximum value of 25

Well lets see.

x-intercepts of 2 and 7 and maximum value of 25

$$y=k(x-2)(x-7)$$

Axis of symmetry will be  x=(2+7)/2 = 4.5

So when x=4.5,  y=25

25=k(4.5-2)(4.5-7)

25=k*2.5*-2.5

25/6.25=-k

k=-4

$$\\y=-4(x-2)(x-7)\\
y=-4(x^2-9x+14)\\
y=-4x^2+36x-56$$

I approached this one a little differently from Melody, but I think we end up with the same solution.

Like her, since x= 2 and x=7 are "zeroes," the x coordinate (because of symmetry) must be at x = 4.5.

And the vertex is given by (4.5, 25)...so.....in "vertex" form, we have...

y = a(x - 4.5)^2 + 25     and we need to find "a"

And we know that when x = 7, y = 0....so we have

0 = a(7 - 4.5)^2 + 25    subtract 25 from both sides

-25 = a(6.25)     divide both sides by 6.25

a = -4       and this should be correct, since the parabola turns "downward"

So, our function is

y = -4(x - 4.5)^2 + 25

Here's the graph.......https://www.desmos.com/calculator/ewlw8wwdni

Yep...that seems correct.....