Question

The graph of the function f(x) = ax2 + bx + c has vertex at (1, 4) and passes through the point (-1, -8). The coefficient a is

Answer 1

the vertex tells us that
f(x) = a(x-1)^2 + 4
using the point (-1,-8) we see that
a(-1-1)^2 + 4 = -8
4a+4 = -8
a = -3
f(x) = -3(x-1)^2 + 4 = -3x^2 + 6x + 1

Answer 2

Vertex has coordinate x = - b / 2a

In this case:

x = 1

x = - b / 2a

1 = - b / 2a

Multiply both sides by 2a

2 a = - b

Multipy both sides by - 1

- 2 a = b

b = - 2 a

Put this value in equation:

f(x) = a x² + b x + c

with coordinates:

x = 1

f = 4

4 = a • 1² + ( - 2 a ) • 1 + c

a - 2 a + c = 4

- a + c = 4

Add a to both sides.

c = a + 4

Again quadratic equation with coordinates:

x = - 1 , f = - 8

x² + b x + c = f(x)

a ( - 1 )² + ( - 2 a ) • ( - 1 ) + a + 4 = - 8

a + 2 a + a + 4 = - 8

4 a + 4 = - 8

Subtract 4 to both sides.

4 a = - 12

a = - 12 / 4

a = - 3

By the way:

b = - 2 a = - 2 • ( - 3 ) = 6

c = a + 4 = - 3 + 4 = 1

Your quadratic equation is:

f(x) = - 3 x² + 6 x + 1