Find the Vertex 5x^2+15x+3

$5 x^{2} + 15 x + 3$

Step 1

Rewrite the equation in vertex form.

Complete the square for $5 x^{2} + 15 x + 3$ .

Use the form $a x^{2} + b x + c$ , to find the values of $a$ , $b$ , and $c$ .

$a = 5$

$b = 15$

$c = 3$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e$

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{15}{2 \cdot 5}$

Cancel the common factor of $15$ and $5$ .

Factor $5$ out of $15$ .

$d = \frac{5 \cdot 3}{2 \cdot 5}$

Cancel the common factors.

Factor $5$ out of $2 \cdot 5$ .

$d = \frac{5 \cdot 3}{5 \cdot 2}$

Cancel the common factor.

$d = \frac{5 \cdot 3}{5 \cdot 2}$

Rewrite the expression.

$d = \frac{3}{2}$

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = 3 - \frac{15^{2}}{4 \cdot 5}$

Simplify the right side.

Simplify each term.

Raise $15$ to the power of $2$ .

$e = 3 - \frac{225}{4 \cdot 5}$

Multiply $4$ by $5$ .

$e = 3 - \frac{225}{20}$

Cancel the common factor of $225$ and $20$ .

Factor $5$ out of $225$ .

$e = 3 - \frac{5 (45)}{20}$

Cancel the common factors.

Factor $5$ out of $20$ .

$e = 3 - \frac{5 \cdot 45}{5 \cdot 4}$

Cancel the common factor.

$e = 3 - \frac{5 \cdot 45}{5 \cdot 4}$

Rewrite the expression.

$e = 3 - \frac{45}{4}$

To write $3$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$e = 3 \cdot \frac{4}{4} - \frac{45}{4}$

Combine $3$ and $\frac{4}{4}$ .

$e = \frac{3 \cdot 4}{4} - \frac{45}{4}$

Combine the numerators over the common denominator.

$e = \frac{3 \cdot 4 - 45}{4}$

Simplify the numerator.

Multiply $3$ by $4$ .

$e = \frac{12 - 45}{4}$

Subtract $45$ from $12$ .

$e = \frac{- 33}{4}$

Move the negative in front of the fraction.

$e = - \frac{33}{4}$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $5 {(x + \frac{3}{2})}^{2} - \frac{33}{4}$ .

$5 {(x + \frac{3}{2})}^{2} - \frac{33}{4}$

Set $y$ equal to the new right side.

$y = 5 {(x + \frac{3}{2})}^{2} - \frac{33}{4}$

Step 2

Use the vertex form, $y = a {(x - h)}^{2} + k$ , to determine the values of $a$ , $h$ , and $k$ .

$a = 5$

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

$k = - \frac{33}{4}$

Step 3

Find the vertex $(h, k)$ .

$(- \frac{3}{2}, - \frac{33}{4})$

Step 4

Find the Vertex 5x^2+15x+3

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