# Question #c26b5

Jun 9, 2017

The equation of parabola is $y = 3 {\left(x - 1\right)}^{2} - 48$

#### Explanation:

Lowest point is vertex of the parabola. y-coordinate of vertex is $- 48$.

x-coordinate of vertex is mid point of x-intercepts $- 3 \mathmr{and} 5$ i.e $1$. So vertex is at $\left(1 , - 48\right)$

The equation of parabola is $y = a {\left(x - h\right)}^{2} + k \mathmr{and} y = a {\left(x - 1\right)}^{2} - 48$.

The point $\left(5 , 0\right)$ is on the parabola , so the point will satisfy the equation. $0 = a {\left(5 - 1\right)}^{2} - 48 \mathmr{and} 0 = 16 a - 48 \mathmr{and} 16 a = 48 \mathmr{and} a = 3$

Hence the equation of parabola is $y = 3 {\left(x - 1\right)}^{2} - 48$.
graph{3(x-1)^2 -48 [-160, 160, -80, 80]} [Ans]