Graph f(x)=x^2+6x+5

Problem

ƒ(x)=x2+6*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 is positive, the parabola opens upward.

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

x=(−6)/(2*(1))=−3

Substitute x=−3 into the function to find the ycoordinate:

ƒ*(−3)=(−3)2+6*(−3)+5=9−18+5=−4

The vertex is (−3,−4)

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

ƒ(0)=0+6*(0)+5=5

The yintercept is (0,5)

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

x2+6*x+5=0

(x+5)*(x+1)=0

The xintercepts are (−5,0) and (−1,0)

5. Plot the points and draw a smooth curve. Use the vertex (−3,−4) the yintercept (0,5) and the xintercepts (−5,0) and (−1,0) to sketch the parabola.

Final Answer

ƒ(x)=(x+3)2−4

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