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

Problem

ƒ(x)=x2+2*x−3

Solution

1. Identify the type of function and its orientation. This is a quadratic function 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=(−2)/(2*(1))=−1

Substitute x=−1 into the function to find the ycoordinate:

ƒ*(−1)=(−1)2+2*(−1)−3=1−2−3=−4

The vertex is (−1,−4)

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

ƒ(0)=0+2*(0)−3=−3

The yintercept is (0,−3)

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

x2+2*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 (−1,−4) the yintercept (0,−3) and the xintercepts (−3,0) and (1,0) to sketch the parabola.

Final Answer

ƒ(x)=(x+1)2−4

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