# How do find the vertex and axis of symmetry, and intercepts for a quadratic equation y= x^2 + 6x + 5?

Mar 30, 2018

Vertex : color(blue)((-3, -4)

Axis of Symmetry is at : color(blue)(x=(-3)

x-intercepts: color(blue)((-1,0) and (-5,0)

y-intercept: color(blue)((0,5)

#### Explanation:

Given:

color(red)(y = f(x) = x^2+6x+5

The Vertex Form of a quadratic function is given by:

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

color(green)(x=h is the axis of symmetry.

Use completing the square method to convert color(red)(f(x) into Vertex Form.

color(red)(y = f(x) = x^2+6x+5

Standard Form $\Rightarrow a {x}^{2} + b x + c = 0$

Consider the quadratic ${x}^{2} + 6 x + 5 = 0$

color(blue)(a=1; b=6 and c=5

Step 1 - Move the constant value to the right-hand side.

Subtract 5 from both sides.

${x}^{2} + 6 x + 5 - 5 = 0 - 5$

${x}^{2} + 6 x + \cancel{5} - \cancel{5} = 0 - 5$

${x}^{2} + 6 x = - 5$

Step 2 - Add a value to both sides.

Add the square of $\frac{b}{2}$

Hence,

${x}^{2} + 6 x + \left[{\left(\frac{6}{2}\right)}^{2}\right] = - 5 + \left[{\left(\frac{6}{2}\right)}^{2}\right]$

${x}^{2} + 6 x + 9 = - 5 + 9$

${x}^{2} + 6 x + 9 = 4$

Step 3 - Write as Perfect Square.

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

Subtract $4$ from both sides to get the vertex form.

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

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

Now, we have the vertex form.

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

Hence, Vertex is at color(blue)((-3,-4)

Axis of Symmetry is at color(red)(x=h

Note that $h = - 3$

rArr color(blue)(x= -3

Step 4 - Write the x, y intercepts.

Consider

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

To find the solutions, take square root on both sides.

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

$\Rightarrow x + 3 = \pm 2$

There are two solutions.

$x + 3 = 2$

$\Rightarrow x = 2 - 3 = - 1$

Hence, $x = - 1$ is one solution.

Next,

$x + 3 = - 2$

$x = - 2 - 3 = - 5$

Hence, $x = - 5$ is the other solution.

Hence, we have two x-intercepts: $\left(- 1 , 0\right) \mathmr{and} \left(- 5 , 0\right)$

To find the y-intercept:

Let $x = 0$

We have,

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

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

$\Rightarrow {3}^{2} - 4 = 9 - 4 = 5$

Hence, y-intercept is at $y = 5$

rArr color(blue)((0,5)

Analyze the image of the graph below: 