Answer:
refer attachment for the graph.
Step-by-step explanation:
Given: The equation [tex]f(x)=x^2-5x+4[/tex]
We have to draw the the graph for the given equation.
Consider the given equation [tex]f(x)=x^2-5x+4[/tex]
The vertex of the parabola of the form [tex]f(x)=ax^2+bx+c[/tex] is given by [tex]x=-\frac{b}{2a}[/tex]
Here, a = 1 , b = -5 and c = 4
Thus, vertex is
[tex]x_v=-\frac{\left(-5\right)}{2\cdot \:1}=\frac{5}{2}[/tex]
Also, the y coordinate at [tex]x=\frac{5}{2}[/tex] is
[tex]y_v=\left(\frac{5}{2}\right)^2-5\cdot \frac{5}{2}+4[/tex]
Simplify, we get,
[tex]y_v=-\frac{9}{4}[/tex]
Thus, The vertex of parabola is [tex]\left(\frac{5}{2},\:-\frac{9}{4}\right)[/tex]
y - intercept is the point where x = 0
Plug x = 0 in given equation [tex]f(x)=x^2-5x+4[/tex]
[tex]f(0)=0^2-5(0)+4=4[/tex]
Thus, y - intercept is (0,4)
Now, we calculate x- intercept
x- intercept is where y is equal to 0.
Put f(x) = 0
We have,
[tex]x^2-5x+4=0[/tex]
Solving the given quadratic equation using quadratic formula ,we have
[tex]x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
we have a = 1 , b = -5 and c = 4
[tex]x_{1,\:2}=\frac{-\left(-5\right)\pm \sqrt{\left(-5\right)^2-4\cdot \:1\cdot \:4}}{2\cdot \:1}[/tex]
Simplify, we have,
[tex]x_{1,2}=\frac{5\pm\sqrt{9}}{2}[/tex]
Thus, [tex]x_1=4, x_2=1[/tex]
Thus, The x - intercept are (4,0) and (0,1)
Plot the graph and we obtain as shown below.
