Graph f(x)=x^2+4x+3

Problem

ƒ(x)=x2+4*x+3

Solution

1. Identify the type of function and its coefficients. This is a quadratic function in the form ƒ(x)=a*x2+b*x+c where a=1 b=4 and c=3 Since a>0 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)+3=4−8+3=−1

The vertex is (−2,−1)

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

ƒ(0)=0+4*(0)+3=3

The yintercept is (0,3)

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

x2+4*x+3=0

(x+3)*(x+1)=0

The xintercepts are (−3,0) and (−1,0)

5. Plot the points and draw a smooth curve. Use the vertex (−2,−1) the yintercept (0,3) and the xintercepts (−3,0) and (−1,0) to sketch the parabola.

Final Answer

ƒ(x)=(x+2)2−1

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