Question

Find the x-intercepts of the parabola with vertex (-7,45) and y-intercept (0,-200). Write your answer in this form: (x1,y1), (x2,V If necessary, round to the nearest hundredth

Answer 1

(-10, 0) , (-4, 0)

Explanation:

`\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}`

`-200=a(0+7)^2+45\\\\\Rightarrow -200=49a+45\\\\\Rightarrow 49a=-245\\\\\Rightarrow a=-(245)/(49)=-5\\\\\text{so the equation of the parabola is}\\\\y=-5(x+7)^2+45`

`\\\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=(-45)/(-5)\\\\\Rightarrow (x+7)^2=9\\\\\Rightarrow (x+7)=\pm √(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)`