How do you find the vertex and intercepts for # f(x) = -3x^2 + 5x + 5#?

1 Answer
Jan 5, 2016

Vertex is at #(5/6,85/12)# Y intercept is at (0,5) and X intercepts are at (-0.703,0) and (2.37,0).

Explanation:

graph{-3x^2+5x+5 [-20, 20, -10, 10]}
This is the equation of Parabola of General form ax^2+bx+c.
Here a=-3 ; b=5 ; c=5 we know the co-ordinate #x_1# of the vertex is equal to (-b/2a). #:.# #x_1 = -5/-6 = 5/6#. Now Putting the value of x -cordinate in the equation we get #y=-3*(5/6)^2+5*5/6+5# or #y=85/12# #:. y_1=85/12# #:.# Vertex is at #(5/6,85/12)#
Now to find Y-intercept putting x=0 we get y = 5 i.e The parabola cuts the Y axis at 5. To get X intercepts putting y=0 ; we get the equation as #-3x^2+5x+5=0# Solving for x in the above quardratic equation we get two roots of x as #-5/(2*(-3))+(sqrt(5^2-4*(-3).5)/(2*(-3)))# and#-5/(2*(-3))-(sqrt(5^2-4*(-3).5)/(2*(-3)))# Which gives the two roots as # -0.703 and 2.37 # So The parabola cuts X-axis at # -0.703 and 2.37 # points. [Answer]