# How to graph a parabola y = (x + 5)^2 - 3?

May 5, 2018

$\text{ }$

#### Explanation:

$\text{ }$
The Vertex Form of a quadratic function is :

color(blue)(y=f(x)=a(x-h)^2+k, where color(green)(( h, k ) is the Vertex of the Parabola.

Quadratic Function is given in Vertex Form: color(red)(y = (x + 5)^2 - 3

color(brown)(h=-5 and k = -3

Vertex is at color(green)((h,k): color(blue)((-5, -3)

Plot the Vertex and the color(red)((x,y) values from the data table.

To find the x-intercepts:

$f \left(x\right) = {\left(x + 5\right)}^{2} - 3$

Let $f \left(x\right) = 0$

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

Add color(red)(3 to both sides:

(x+5)^2-3+color(red)(3)= 0+color(red)(3

(x+5)^2-cancel 3+color(red)(cancel 3)= 0+color(red)(3

${\left(x + 5\right)}^{2} = 3$

Take Square root on both sides to simplify:

$\sqrt{{\left(x + 5\right)}^{2}} = \sqrt{3}$

$\left(x + 5\right) = \pm \sqrt{3}$

Subtract $\textcolor{red}{5}$ from both sides:

$\left(x + 5\right) - \textcolor{red}{5} = \pm \sqrt{3} - \textcolor{red}{5}$

$\left(x + \cancel{5}\right) - \textcolor{red}{\cancel{5}} = \pm \sqrt{3} - \textcolor{red}{5}$

$x = \pm \sqrt{3} - 5$

Hence, color(blue)(x=[sqrt(3)-5] is one solution and color(blue)(x=[-sqrt(3)-5] is the other.

Using a calculator,

$\textcolor{b l u e}{x \approx - 3.26795}$ is one solution.

$\textcolor{b l u e}{x \approx - 6.73205}$ is another solution.

Hence, x-intercepts are: $x \approx - 3.3$, $x \approx - 6.7$

Verify this solution by using graphs:

color(green)("Graph 1"

Graph of color(blue)(y=x^2

This is the Parent Graph.

Use this graph to understand the behavior of the given quadratic function.

color(green)("Graph 2"

Graph of color(blue)(y = (x + 5)^2 - 3

Study the graphs of both the Parent function and the given function.

Next, verify the x-intercepts:

Hope it helps.