# How do you find the vertex and intercepts for y = (x + 3)^2 – 4?

Dec 29, 2017

Vertex $\to \left(x , y\right) = \left(- 3 , - 4\right)$
${x}_{\text{intercepts}} \to \left(x , y\right) = \left(- 5 , 0\right) \mathmr{and} \left(- 1 , 0\right)$
${y}_{\text{intercept}} \to \left(x , y\right) = \left(0 , 5\right)$

#### Explanation:

$\textcolor{b l u e}{\text{Determine the "x" intercepts}}$

Set $y = 0 = {\left(x + 3\right)}^{2} - 4$

$4 = {\left(x + 3\right)}^{2}$

Square root both sides

$\pm 2 = x + 3$

Subtract 3 from both sides

$x = - 3 \pm 2$

${x}_{\text{intercepts}} \to \left(x , y\right) = \left(- 5 , 0\right) \mathmr{and} \left(- 1 , 0\right)$

You may if you so choose determine the vertex from this point

${x}_{\text{vertex}}$ is midway between the x intercepts then by substitution determine y. The method I used later takes less work.

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$\textcolor{b l u e}{\text{Determin "y" intercept and vertex}}$

Given $y {\left(x \textcolor{m a \ge n t a}{+ 3}\right)}^{2} \textcolor{g r e e n}{- 4}$

${x}_{\text{vertex}} = \left(- 1\right) \times \left(\textcolor{m a \ge n t a}{+ 3}\right) = - 3$
${y}_{\text{vertex}} = \textcolor{g r e e n}{- 4}$

Vertex $\to \left(x , y\right) = \left(- 3 , - 4\right)$

${y}_{\text{intercept}} = {\textcolor{m a \ge n t a}{+ 3}}^{2} \textcolor{g r e e n}{- 4} = 5$

Dec 29, 2017

Vertex is at $\left(- 3 , - 4\right)$ , y intercept is $y = 5 \mathmr{and} \left(0 , 5\right)$
x intercepts are at $\left(- 1 , 0\right) \mathmr{and} \left(- 5 , 0\right)$

#### Explanation:

$y = {\left(x + 3\right)}^{2} - 4$

Comparing with vertex form of equation

f(x) = a(x-h)^2+k ; (h,k) being vertex we find

here $h = - 3 , k = - 4 \therefore$ Vertex is at $\left(- 3 , - 4\right)$

y intercept is found by putting $x = 0$ in the equation

$y = {\left(x + 3\right)}^{2} - 4 \therefore y = {\left(0 + 3\right)}^{2} - 4 = 9 - 4 = 5$

So y intercept is $y = 5 \mathmr{and} \left(0 , 5\right)$

x intercepts are found by putting $y = 0$ in the equation

$y = {\left(x + 3\right)}^{2} - 4 \therefore 0 = {\left(x + 3\right)}^{2} - 4$ or

${\left(x + 3\right)}^{2} = 4 \mathmr{and} \left(x + 3\right) = \pm \sqrt{4} \mathmr{and} \left(x + 3\right) = \pm 2$

$\therefore x = - 3 \pm 2 \therefore x = - 1 \mathmr{and} x = - 5$

x intercepts are at $\left(- 1 , 0\right) \mathmr{and} \left(- 5 , 0\right)$

graph{(x+3)^2-4 [-10, 10, -5, 5]} [Ans]

Dec 29, 2017

Vertex = $\left(- 3 , - 4\right)$

y-intercept = $\left(0 , 5\right)$

x-intercepts = $\left(- 1 , 0\right)$ and $\left(- 5 , 0\right)$

#### Explanation:

First, expand $y = {\left(x + 3\right)}^{2} - 4$ into $y = {x}^{2} + 6 x + 5$.

The x-vertex of a quadratic equation $y = a {x}^{2} + b x + c$ is given by $- \frac{b}{2} a$.

Plugging in $a = 1$, $b = 6$, and $c = 5$, we find that the x-vertex is $- \frac{6}{2} = - 3$.

Now we plug in $x = - 3$ into $y = {x}^{2} + 6 x + 5$ to get the y-vertex, which is $- 4$.

So our vertex is $\left(- 3 , - 4\right)$.

To find the y-intercept of a quadratic equation, plug in $0$ for the x value, and you will get $5$.

To find the x-intercept of a quadratic equation, plug in $0$ for the y value, and you will get two solutions, $x = - 1$ and $x = - 5$.

So our y-intercept is $\left(0 , 5\right)$, and our x-intercepts are $\left(- 1 , 0\right)$ and $\left(- 5 , 0\right)$.