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