Given
[tex]y=x^2-4x+3[/tex]
Find
Graph the parabola of the given function
Explanation
[tex]y=x^2-4x+3[/tex]
solve the equation
[tex]\begin{gathered} x^2-4x+3=0 \\ x^2-3x-x+3=0 \\ x(x-3)-(x-3)=0 \\ (x-1)(x-3)=0 \\ x=1,3 \end{gathered}[/tex]
vertex can be found by using the formula,
[tex]-\frac{b}{2a}=-\frac{-4}{2}=2[/tex]
x = 2 , substitute this in equation to get y value,
y = -1
if x = 0 then y =3 and if y= 0 then x = 1, 3
Final Answer
