Find a such that f(x)= ax^2-3x+5 has a maximum value of 15.

Question

I'm stuck because how am I suppose to complete the square? 3/2= 1.5? then square 1.5? So lost.

Answer 1

well, even without completing the square, you know the vertex is at x = -b/2a = 3/2a

So,
a(3/(2a))^2 - 3(3/(2a)) + 5 = 15
5 - 9/(4a) = 15
a = -9/40

But, in the spirit of completing the square,

ax^2-3x+5
= a(x^2 - 3/a) + 5
= a(x^2 - 3/a + (3/2a)^2) + 5 - a(3/2a)^2
= a(x - 3/2a)^2 + 5 - 9/4a

This is the vertex form
y = a(x-h)^2 + k
so we have

5 - 9/4a = 15
9/4a = -10
4a = -9/10
a = -9/40

Answer 2

the max is on the axis of symmetry

x = 3 / 2a

15 = a (3 / 2a)^2 - 3 (3 / 2a) + 5

10 = (9 / 4a) - (9 / 2a)

10 a = 9/4 - 9/2 = - 9/4

a = -9/40