# How do you find the x-intercepts of the parabola with vertex (4,-1) and y-intercept of (0,15)?

"x"-intercepts of the parabola are $\left(3 , 0\right) \mathmr{and} \left(5 , 0\right)$
The equation of parabola with vertex $\left(h , k\right) \mathmr{and} \left(4 , - 1\right)$ is $y = a {\left(x - h\right)}^{2} + k \mathmr{and} y = a {\left(x - 4\right)}^{2} - 1$. The parabola passes through point $\left(0 , 15\right)$. So the point will satisfy the equation of parabola.
$15 = a {\left(0 - 4\right)}^{2} - 1 \mathmr{and} 15 = 16 a - 1 \mathmr{and} 16 a = 16 \mathmr{and} a = 1$
Hence the equation of parabola is $y = {\left(x - 4\right)}^{2} - 1$. To find x -intercepts putting $y = 0$ in the equation we get $0 = {\left(x - 4\right)}^{2} - 1 \mathmr{and} {x}^{2} - 8 x + 15 = 0 \mathmr{and} \left(x - 3\right) \left(x - 5\right) = 0 \therefore x = 3 \mathmr{and} x = 5$. So x-intercepts are $\left(3 , 0\right) \mathmr{and} \left(5 , 0\right)$ graph{(x-4)^2-1 [-11.25, 11.25, -5.625, 5.625]}[Ans]