Question 417420: find the x-intercept of a parabola with a vertex (-1,2) and y-intercept (0,-3)
Answer by Gogonati(855)   (Show Source):
You can put this solution on YOUR website! The equation of parabola is;(x-x')^2=4c(y-y') where (x',y') is a point on the graph. Since (-1,2) is the vertex we write;(x+1)^2=4c(y-2) where y=-c is the directrix of parabola Since the y-intercept is (0,-3) we have; (0+1)^2=4c(-3-2)
after solving this equation we find c=-1/20 Thus the equation of the parabola will be; (x+1)^2=-1/5(y-2) or -5x^2-10x-3=y To find the x-intercept make y=0 and solve the equation; 5x^2+10x+3=0 and have two points; [-1+sqrt(40)/10, 0]and
[-1-sqrt(40)/10] who parabola intersect the x-axis.
graph( 600,400, -10, 10, -10, 10, 5x^2+10x+3 )
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