Graph y=x^2-2x-3

Problem

y=x2−2*x−3

Solution

1. Identify the type of function and its orientation. This 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))

x=1

Substitute x=1 into the original equation to find the ycoordinate.

y=(1)2−2*(1)−3

y=−4

The vertex is (1,−4)

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

y=(0)2−2*(0)−3

y=−3

The yintercept is (0,−3)

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

0=x2−2*x−3

0=(x−3)*(x+1)

Setting each factor to zero gives x=3 and x=−1 The xintercepts are (3,0) and (−1,0)

5. Plot the points and draw a smooth curve. Use the vertex (1,−4) the intercepts (0,−3) (3,0) (−1,0) and the axis of symmetry x=1 to sketch the parabola.

Final Answer

y=x2−2*x−3

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

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