We are given a quadratic function and its vertex, and want to find the values of

a

and

b .

\underline{Function} y = a x^{2} + b x + 8 \underline{Vertex} (2, - 4)

To do so, we will use the vertex form of a quadratic function.

y = a (x - h)^{2} + k

In this form, the vertex is

(h, k) .

Since we know the vertex of our function is

(2, - 4),

we can substitute

2

for

h

and

- 4

for

k .

y = a (x - 2)^{2} + (- 4) ⇕ y = a (x - 2)^{2} - 4

Let's simplify this equation.

y = a (x - 2)^{2} - 4

$(a - b)^{2} = a^{2} - 2 a b + b^{2}$

y = a (x^{2} - 2 (x) (2) + 2^{2}) - 4

Simplify power and product

y = a (x^{2} - 4 x + 4) - 4

Distribute $a$

y = a x^{2} - 4 a x + 4 a - 4

Now, we can compare the given function with the one we obtained, term by term.

We can find the value of

a

by equating the constant terms.

8 = 4 a - 4

▼

Simplify

$LHS + 4 = RHS + 4$

12 = 4 a

$LHS / 4 = RHS / 4$

3 = a

Rearrange equation

a = 3

Finally, from the linear coefficient, we know that

b = - 4 a .

We will find the value of

b

by substituting

3

for

a .

b = - 4 (3) \Leftrightarrow b = - 12

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