Graph f(x)=x^2-6x+8

Problem

ƒ(x)=x2−6*x+8

Solution

1. Identify the coefficients of the quadratic function in the form ƒ(x)=a*x2+b*x+c

a=1

b=−6

c=8

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

x=(−(−6))/(2*(1))

x=3

3. Find the y-coordinate of the vertex by evaluating ƒ(3)

ƒ(3)=(3)2−6*(3)+8

ƒ(3)=9−18+8

ƒ(3)=−1

The vertex is (3,−1)

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

ƒ(0)=(0)2−6*(0)+8

ƒ(0)=8

The y-intercept is (0,8)

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

x2−6*x+8=0

(x−2)*(x−4)=0

x=2,x=4

The x-intercepts are (2,0) and (4,0)

6. Sketch the parabola using the vertex (3,−1) the intercepts, and the fact that it opens upward since a>0

Final Answer

ƒ(x)=(x−3)2−1

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