# How do you find the vertex and axis of symmetry, and then graph the parabola given by: y=x^2 +2x -5?

Jul 17, 2018

Vertex is at $\left(- 1 , - 6.\right)$ , axis of symmetry is $x = - 1$
y-intercept is at $\left(0 , - 5\right)$, x-intercepts are at $\left(- 3.45 , 0\right) \mathmr{and} \left(1.45 , 0\right)$

#### Explanation:

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

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

$y = {\left(x + 1\right)}^{2} - 6$ . This is vertex form of equation

y=a(x-h)^2+k ; (h,k) being vertex ,

:.h=-1 ,k=-6,a=1 ; a>0 :.Parabola opens upward.

Therefore vertex is at $\left(- 1 , - 6.\right)$ . Axis of symmetry is

$x = h \mathmr{and} x = - 1$ . Graphing: y-intercept is at

$x = 0 \therefore y = {x}^{2} + 2 x - 5 = - 5 \mathmr{and} \left(0 , - 5\right)$, x-intercepts are

at y=0 ; y=(x+1)^2-6 or (x+1)^2-6=0  or

${\left(x + 1\right)}^{2} - 6 = 0 \mathmr{and} {\left(x + 1\right)}^{2} = 6 \mathmr{and} \left(x + 1\right) = \pm \sqrt{6}$ or

$x = - 1 \pm \sqrt{6} \mathmr{and} x \approx 1.45 , x = - 3.45$ Therefore,

x-intercepts are at $\left(- 3.45 , 0\right) \mathmr{and} \left(1.45 , 0\right)$

graph{x^2+2x-5 [-20, 20, -10, 10]} [Ans]