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 \underline{Vertex} (- 3, 2)

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

(- 3, 2),

we can substitute

- 3

for

h

and

2

for

k .

y = a (x - (- 3))^{2} + 2 ⇕ y = a (x + 3)^{2} + 2

Let's simplify this equation.

y = a (x + 3)^{2} + 2

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

y = a (x^{2} + 2 (x) (3) + 3^{2}) + 2

Simplify power and product

y = a (x^{2} + 6 x + 9) + 2

Distribute $a$

y = a x^{2} + 6 a x + 9 a + 2

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.

0 = 9 a + 2

▼

Simplify

$LHS - 2 = RHS - 2$

- 2 = 9 a

$LHS / 9 = RHS / 9$

\frac{- 2}{9} = a

Put minus sign in front of fraction

- \frac{2}{9} = a

Rearrange equation

a = - \frac{2}{9}

Finally, from the linear coefficient, we know that

b = 6 a .

We will find the value of

b

by substituting

- \frac{2}{9}

for

a .

b = 6 a

$a = - \frac{2}{9}$

b = 6 (- \frac{2}{9})

▼

Simplify right-hand side

$a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$

b = - \frac{1 2}{9}

$\frac{a}{b} = \frac{a / 3}{b / 3}$

b = - \frac{4}{3}

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