How do you find the vertex and intercepts for f(x)=-5x^2+3x+8f(x)=−5x2+3x+8?

1 Answer
Nghi N.
Nov 17, 2015

Find vertex and intercepts of f(x) = -5x^2 + 3x + 8f(x)=−5x2+3x+8

Explanation:

x-coordinate of vertex:
x = -b/(2a) = -3/-10 = 3/10.x=−b2a=−3−10=310.
y-coordinate of vertex:
f(3/10) = - 5(9/100) + 3(3/10) + 8 = = -45/100 + 90/100 + 800/100 = 845/100 = 8.45f(310)=−5(9100)+3(310)+8==−45100+90100+800100=845100=8.45
To find y-intercept, make x = 0 --> y = 8
To find x-intercepts, make y = 0. Solve the quadratic equation:
- 5x^2 + 3x + 8 = 0.−5x2+3x+8=0.
Since a - b + c = 0, use shortcut: the 2 real roots are: x = -1x=−1 and
x = -c/a = 8/5x=−ca=85

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