Graph f(x)=4x^2-x-3

Problem

ƒ(x)=4*x2−x−3

Solution

1. Identify the type of function and its orientation. The function is a quadratic of the form ƒ(x)=a*x2+b*x+c where a=4 b=−1 and c=−3 Since a>0 the parabola opens upward.

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

x=(−(−1))/(2*(4))

x=1/8

Substitute x=1/8 into the function to find the ycoordinate.

ƒ(1/8)=4*(1/8)2−1/8−3

ƒ(1/8)=4/64−1/8−3

ƒ(1/8)=1/16−2/16−48/16

ƒ(1/8)=−49/16

The vertex is (1/8,−3.0625)

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

ƒ(0)=4*(0)2−0−3

ƒ(0)=−3

The yintercept is (0,−3)

4. Find the x-intercepts by setting ƒ(x)=0 and solving the quadratic equation.

4*x2−x−3=0

(4*x+3)*(x−1)=0

4*x+3=0⇒x=−3/4

x−1=0⇒x=1

The xintercepts are (−3/4,0) and (1,0)

5. Plot the points and draw the smooth U-shaped curve (parabola) passing through the vertex (1/8,−49/16) the yintercept (0,−3) and the xintercepts (−3/4,0) and (1,0)

Final Answer

ƒ(x)=4*x2−x−3

Want more problems? Check here!