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

Jul 11, 2015

The vertex is at ($- 1 , - 2$).
The axis of symmetry is $x = 1$.
The $y$-intercept is at ($0 , 5$).
The $x$-intercepts are at $x = - 1 - \sqrt{6}$ and $x = - 1 + \sqrt{6}$.

#### Explanation:

The standard form for the equation of a parabola is

$y = {a}^{2} + b x + c$

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

So

$a = 1$, $b = 2$, and $c = 5$.

Vertex

Since $a > 0$, the parabola opens upwards.

The $x$-coordinate of the vertex is at

x = –b/(2a) = -2/(2×1) = -2/2 = -1.

Insert this value of $x$ back into the equation.

$y = {x}^{2} + 2 x - 5 = {\left(- 1\right)}^{2} + 2 \left(- 1\right) - 5 = 1 - 2 - 5 = - 6$

The vertex is at ($- 1 , - 2$).

Axis of symmetry

The axis of symmetry must pass through the vertex, so

The axis of symmetry is $x = - 1$.

$y$-intercept

To find the $y$-intercept, we set $x = 0$ and solve for $y$.

y = x^2 + 2x – 5 = 0^2 + 2×0 -5 = 0 + 0 – 5 = -5

The $y$-intercept is at ($0 , 5$).

$x$-intercepts

To find the $x$-intercepts, we set $y = 0$ and solve for $x$.

y = x^2 + 2x – 5

0 = x^2 + 2x – 5

x = (-b ±sqrt(b^2-4ac))/(2a)

x = (-2 ± sqrt(2^2 - 4×1×(-5)))/(2×1)

x = (-2 ± sqrt(2^2 - 4×1×(-5)))/(2×1)

x = (-2 ± sqrt(4 + 20))/2

x = (-2 ± sqrt24)/2

x = -1 ± 1/2sqrt24

x = -1 ± 1/2(sqrt(4×6))

x = -1 ± 1/2(2sqrt6)

x = -1 ± sqrt6

The $x$-intercepts are at $x = - 1 - \sqrt{6}$ and $x = - 1 + \sqrt{6}$.

Graph

Now we prepare a chart.

The axis of symmetry passes through $x = - 1$.

Let's prepare a table with points that are 5 units on either side of the axis, that is, from $x = - 6$ to $x = 4$. Plot these points. And we have our graph.