Graph f(x)=x^2+4x-5

Problem

ƒ(x)=x2+4*x−5

Solution

1. Identify the type of function and its orientation. The function is a quadratic in the form ƒ(x)=a*x2+b*x+c Since a=1 (which is positive), the parabola opens upward.

2. Find the vertex using the formula x=(−b)/(2*a)

x=(−4)/(2*(1))=−2

Substitute x=−2 into the function to find the ycoordinate:

ƒ*(−2)=(−2)2+4*(−2)−5=4−8−5=−9

The vertex is (−2,−9)

3. Determine the y-intercept by evaluating ƒ(0)

ƒ(0)=0+4*(0)−5=−5

The yintercept is (0,−5)

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

x2+4*x−5=0

(x+5)*(x−1)=0

The xintercepts are x=−5 and x=1 or the points (−5,0) and (1,0)

5. Plot the points and draw the parabola. Use the vertex (−2,−9) the yintercept (0,−5) and the xintercepts (−5,0) and (1,0) to sketch the curve.

Final Answer

ƒ(x)=(x+2)2−9

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