Question

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

Answer 1

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]