Graph y=x^2-4x+3

Problem

y=x2−4*x+3

Solution

1. Identify the type of function and its orientation. The equation is a quadratic function in the form y=a*x2+b*x+c Since a=1 which is positive, the parabola opens upward.

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

x=(−(−4))/(2*(1))

x=2

3. Calculate the y-coordinate of the vertex by substituting x=2 into the original equation.

y=(2)2−4*(2)+3

y=4−8+3

y=−1

The vertex is (2,−1)

4. Find the y-intercept by setting x=0

y=0−4*(0)+3

y=3

The y-intercept is (0,3)

5. Find the x-intercepts by setting y=0 and factoring the quadratic.

x2−4*x+3=0

(x−1)*(x−3)=0

x=1

x=3

The x-intercepts are (1,0) and (3,0)

6. Plot the points and draw a smooth curve. Use the vertex (2,−1) the intercepts (0,3) (1,0) (3,0) and the point of symmetry (4,3) to sketch the parabola.

Final Answer

y=x2−4*x+3

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