Answer:
(-10, 0) , (-4, 0)
Step-by-step explanation:
[tex]\text{We know that the parabola is given by the quadratic equation.}\\\text{let the general equation of the parabola is}\\\\y=a(x-h)^2+k, \text{ where (h, k) is the vertex of the parabola.}\\\\\text{here we have given that vertex, }(h,k)=(-7,45), \text{ so above equation gives}\\\\y=a(x-(-7))^2+(45)\\\\y=a(x+7)^2+45\\\\\text{now this parabola has y-intercept at }(0,-200), \text{ so with this point}\\\text{above equation gives}[/tex]
[tex]-200=a(0+7)^2+45\\\\\Rightarrow -200=49a+45\\\\\Rightarrow 49a=-245\\\\\Rightarrow a=-\frac{245}{49}=-5\\\\\text{so the equation of the parabola is}\\\\y=-5(x+7)^2+45[/tex]
[tex]\\\text{Now to find the x-intercepts, we set }y=0, \text{ so we get}\\\\-5(x+7)^2+45=0\\\\\Rightarrow -5(x+7)^2=-45\\\\\Rightarrow (x+7)^2=\frac{-45}{-5}\\\\\Rightarrow (x+7)^2=9\\\\\Rightarrow (x+7)=\pm \sqrt{9}\\\\\Rightarrow x+7=\pm 3\\\\\Rightarrow x=-7\pm 3\\\\\Rightarrow x=-7-3, \text{ and } x=-7+3\\\\\Rightarrow x=-10, \text{ and } x=-4\\\\\text{so x-intercepts of parabola occur at: }(-10,0), \text{ and }(-4,0)[/tex]
