Graph y=x^2+2x-3

Problem

y=x2+2*x−3

Solution

1. Identify the type of function and its orientation. The equation is a quadratic function in the form y=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=−2/(2*(1))=−1

Substitute x=−1 into the original equation to find the ycoordinate:

y=(−1)2+2*(−1)−3=1−2−3=−4

The vertex is (−1,−4)

3. Find the y-intercept by setting x=0

y=(0)2+2*(0)−3=−3

The yintercept is (0,−3)

4. Find the x-intercepts by setting y=0 and factoring the quadratic.

x2+2*x−3=0

(x+3)*(x−1)=0

The xintercepts are x=−3 and x=1 or the points (−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

y=x2+2*x−3

The graph is an upward-opening parabola with vertex (−1,−4) yintercept (0,−3) and xintercepts (−3,0) and (1,0)

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