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

Problem

ƒ(x)=x2−2*x−3

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 where a=1 b=−2 and c=−3 Since a>0 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=−4

The vertex is (1,−4)

3. Determine the y-intercept by evaluating the function at x=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 the quadratic.

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 (−1,0) and (3,0) to sketch the parabola.

Final Answer

ƒ(x)=(x−1)2−4

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