# How do you graph the parabola x^2 + 10x +8y +17 = 0 using vertex, intercepts and additional points?

Jul 21, 2018

Vertex: $\left(- 5 , 1\right)$, symmetry: $x = - 5$, y intercept: $\left(0 , - 2.125\right)$, x intercepts: $\left(- 7.83 , 0\right) \mathmr{and} \left(- 2.17 , 0\right)$, additional points : $\left(3 , - 7\right) \mathmr{and} - 13 , - 7$

#### Explanation:

${x}^{2} + 10 x + 8 y + 17 = 0 \mathmr{and} 8 y = - \left({x}^{2} + 10 x\right) - 17$ or

$y = - \frac{1}{8} \left({x}^{2} + 10 x\right) - \frac{17}{8}$ or

$y = - \frac{1}{8} \left({x}^{2} + 10 x + 25\right) + \frac{25}{8} - \frac{17}{8}$ or

$y = - \frac{1}{8} {\left(x + 5\right)}^{2} + 1$ This is vertex form of equation ,

y=a(x-h)^2+k ;(h,k) is vertex , $h = - 5 , k = 1 , a = - \frac{1}{8}$

Since $a$ is negative, parabola opens downward.

Therefore vertex is at $\left(- 5 , 1\right)$ Axis of symmetry is

x= h or x = -5 ;  y-intercept is found by putting $x = 0$

in the equation ${0}^{2} + 10 \cdot 0 + 8 y + 17 = 0 \mathmr{and} 8 y = - 17$ or

$y = - \frac{17}{8} = - 2.125$ or at $\left(0 , - 2.125\right)$

x-intercepts are found by putting $y = 0$

in the equation ${x}^{2} + 10 x + 17 = - 8 y \mathmr{and} {x}^{2} + 10 x = - 17$ or

${x}^{2} + 10 x + 25 = 25 - 17 \mathmr{and} {\left(x + 5\right)}^{2} = 8 \mathmr{and} x + 5 = \pm \sqrt{8}$

$\therefore x = - 5 \pm \sqrt{8} \mathmr{and} x \approx - 7.83 , x \approx - 2.17$ , therefore ,

x-intercepts are at $\left(- 7.83 , 0\right) \mathmr{and} \left(- 2.17 , 0\right)$

Additional points: $x = 3 , y = - \frac{1}{8} {\left(3 + 5\right)}^{2} + 1$ or

$y = - \frac{1}{8} \cdot {8}^{2} + 1 \mathmr{and} y = - 7 \therefore \left(3 , - 7\right)$

$x = - 13 , y = - \frac{1}{8} {\left(- 13 + 5\right)}^{2} + 1$ or

$y = - \frac{1}{8} \cdot {\left(- 8\right)}^{2} + 1 \mathmr{and} y = - 7 \therefore \left(- 13 , - 7\right)$

graph{x^2+10x+8y+17=0 [-20, 20, -10, 10]}[Ans]