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

Question

Eli Assouline · Accepted Answer

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).

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]

Ethan Adkins · Answer

undefined To find the vertex and intercepts for the quadratic function $ f(x) = -3x^2 + 5x + 5 $, follow these steps:

1. **Vertex**: To find the vertex of the quadratic function, use the formula $ x = -\frac{b}{2a} $, where $ a $ and $ b $ are the coefficients of the quadratic term and the linear term, respectively. Once you find the value of $ x $, substitute it back into the original function to find the corresponding value of $ y $.

$ x = -\frac{5}{2(-3)} = \frac{5}{6} $

Now, substitute $ x = \frac{5}{6} $ into the function:
   $ f\left(\frac{5}{6}\right) = -3\left(\frac{5}{6}\right)^2 + 5\left(\frac{5}{6}\right) + 5 = -\frac{25}{4} + \frac{25}{6} + 5 = \frac{5}{12} $

So, the vertex of the quadratic function is $ \left(\frac{5}{6}, \frac{5}{12}\right) $.

2. **Intercepts**:
   - **x-intercepts**: Set $ f(x) = 0 $ and solve for $ x $.
     $ -3x^2 + 5x + 5 = 0 $

You can use the quadratic formula or factorization to find the roots. Let's use the quadratic formula:
     $ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $

$ x = \frac{-5 \pm \sqrt{5^2 - 4(-3)(5)}}{2(-3)} $
     $ x = \frac{-5 \pm \sqrt{25 + 60}}{-6} $
     $ x = \frac{-5 \pm \sqrt{85}}{-6} $

Therefore, the x-intercepts are $ x = \frac{-5 + \sqrt{85}}{-6} $ and $ x = \frac{-5 - \sqrt{85}}{-6} $.

- **y-intercept**: To find the y-intercept, set $ x = 0 $ and solve for $ y $.
     $ f(0) = -3(0)^2 + 5(0) + 5 = 5 $

So, the y-intercept is $ (0, 5) $.