Graph f(x)=2x^2-x-1

Problem

ƒ(x)=2*x2−x−1

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=2 which is positive, the parabola opens upward.

2. Find the vertex using the formula x=(−b)/(2*a)

x=(−(−1))/(2*(2))

x=1/4

Substitute x=1/4 into the function to find the ycoordinate:

ƒ(1/4)=2*(1/4)2−1/4−1

ƒ(1/4)=2/16−4/16−16/16

ƒ(1/4)=−18/16=−1.125

The vertex is (1/4,−1.125)

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

ƒ(0)=2*(0)2−0−1

ƒ(0)=−1

The yintercept is (0,−1)

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

2*x2−x−1=0

(2*x+1)*(x−1)=0

x=−1/2

x=1

The xintercepts are (−1/2,0) and (1,0)

5. Plot the points and draw a smooth U-shaped curve through the vertex, the yintercept, and the xintercepts.

Final Answer

ƒ(x)=2*x2−x−1

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