Graph f(x)=x^2+2x-8

Problem

ƒ(x)=x2+2*x−8

Solution

1. Identify the vertex by using the formula x=(−b)/(2*a) where a=1 and b=2

x=(−2)/(2*(1))=−1

2. Calculate the y-coordinate of the vertex by substituting x=−1 into the function.

ƒ*(−1)=(−1)2+2*(−1)−8=−9

The vertex is at (−1,−9)

3. Find the y-intercept by evaluating the function at x=0

ƒ(0)=0+2*(0)−8=−8

The y-intercept is at (0,−8)

4. Find the x-intercepts by setting ƒ(x)=0 and factoring the quadratic expression.

x2+2*x−8=0

(x+4)*(x−2)=0

The x-intercepts are at x=−4 and x=2

5. Determine the orientation of the parabola. Since the leading coefficient a=1 is positive, the parabola opens upward.

Final Answer

ƒ(x)=(x+1)2−9

The graph is a parabola opening upward with vertex (−1,−9) y-intercept (0,−8) and x-intercepts (−4,0) and (2,0)

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