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

Problem

ƒ(x)=x2−4*x

Solution

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

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

x=(−(−4))/(2*(1))

x=2

Substitute x=2 into the function to find the ycoordinate:

ƒ(2)=(2)2−4*(2)

ƒ(2)=−4

The vertex is (2,−4)

3. Find the x-intercepts by setting ƒ(x)=0 and factoring.

x2−4*x=0

x*(x−4)=0

x=0

x=4

The xintercepts are (0,0) and (4,0)

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

ƒ(0)=(0)2−4*(0)

ƒ(0)=0

The yintercept is (0,0)

5. Plot the vertex (2,−4) and the intercepts (0,0) and (4,0) then draw a smooth U-shaped curve through these points.

Final Answer

ƒ(x)=x2−4*x

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