We have a quadratic function written in standard form, and we want to rewrite it in vertex form.

\underline{Standard Form} y = a x^{2} + b x + c \underline{Given Equation} y = \frac{9}{4} x^{2} + 3 x - 1 \Leftrightarrow y = \frac{9}{4} x^{2} + 3 x + (- 1)

In the given equation,

a = \frac{9}{4},

b = 3,

and

c = - 1 .

Let's now recall the vertex form of a quadratic function.

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

In this equation,

a

is the leading coefficient of the quadratic function, and the point

(h, k)

is the vertex of the parabola. By substituting our given values for

a

and

b

into the expression

- \frac{b}{2 a},

we can find

h .

- \frac{b}{2 a}

$a = \frac{9}{4}$ , $b = 3$

- \frac{3}{2 ( \frac{9}{4} )}

▼

Simplify

$\frac{a}{b} = \frac{a \cdot 2}{b \cdot 2}$

- \frac{6}{4 \cdot \frac{9}{4}}

$4 \cdot \frac{a}{4} = a$

- \frac{6}{9}

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

- \frac{2}{3}

So far, we know that the vertex lies at

(- \frac{2}{3}, k) .

To find the

y -

coordinate

k,

we will substitute

- \frac{2}{3}

for

x

in the given function.

y = \frac{9}{4} x^{2} + 3 x - 1

$x = - \frac{2}{3}$

y = \frac{9}{4} (- \frac{2}{3})^{2} + 3 (- \frac{2}{3}) - 1

▼

Simplify right-hand side

$(\frac{a}{b})^{m} = \frac{a ^{m}}{b ^{m}}$

y = \frac{9}{4} (\frac{4}{9}) + 3 (- \frac{2}{3}) - 1

$\frac{a}{b} \cdot \frac{b}{a} = 1$

y = 1 + 3 (- \frac{2}{3}) - 1

$3 \cdot \frac{a}{3} = a$

y = 1 - 2 - 1

Subtract terms

y = - 2

Therefore, the

(h, k)

coordinate pair of the vertex is

(- \frac{2}{3}, - 2) .

Moreover, since we already know that

a = \frac{9}{4},

we can rewrite the given function in vertex form.

y = \frac{9}{4} (x - (- \frac{2}{3}))^{2} + (- 2) ⇕ y = \frac{9}{4} (x + \frac{2}{3})^{2} - 2

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