we have the function
[tex]f\left(x\right)=x^2-5x+4[/tex]
This is a vertical parabola open upward
The vertex is a minimum
step 1
Find out the vertex
Complete the square
[tex]\begin{gathered} f(x)=(x^2-5x)+4 \\ f(x)=(x^2-5x+2.5^2-2.5^2)+4 \\ f(x)=(x-2.5)^2-2.25 \end{gathered}[/tex]
The vertex is the point (2.5,-2.25)
step 2
Find out the y-intercept (value of f(x) when the value of x=0)
so
For x=0
[tex]\begin{gathered} f(x)=0^2-5(0)+4 \\ f(x)=4 \end{gathered}[/tex]
The y-intercept is (0,4)
step 3
Find out the x-intercepts (values of x when the value of f(x)=0)
[tex]x^2-5x+4=0[/tex]
Solve the quadratic equation
using the formula
a=1
b=-5
c=4
substitute in the formula
[tex]x=\frac{-(-5)(+-)\sqrt{-5^2-4(1)(4)}}{2(1)}[/tex][tex]x=\frac{5(+-)+3}{2}[/tex]
The values of x are
x=4 and x=1
therefore
the x-intercepts are (1,0) and (4,0)
therefore
the graph of the given equation is
