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

Problem

ƒ(x)=x2−4*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=−4 and c=3 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)+3

ƒ(2)=4−8+3

ƒ(2)=−1

The vertex is (2,−1)

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

ƒ(0)=(0)2−4*(0)+3

ƒ(0)=3

The yintercept is (0,3)

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

x2−4*x+3=0

(x−1)*(x−3)=0

x=1

x=3

The xintercepts are (1,0) and (3,0)

5. Plot the points and draw the smooth curve of the parabola through the vertex (2,−1) the yintercept (0,3) and the xintercepts (1,0) and (3,0)

Final Answer

ƒ(x)=(x−2)2−1

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