Question

Which of the following is the graph of f(x) = x2 − 5x + 4?

Answer 1

I made a graph of the equation hope this helps.

Answer 2

refer attachment for the graph.

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=-(b)/(2a)`

Here, a = 1 , b = -5 and c = 4

Thus, vertex is

`x_v=-(\left(-5\right))/(2\cdot \:1)=(5)/(2)`

Also, the y coordinate at
`x=(5)/(2)` is

`y_v=\left((5)/(2)\right)^2-5\cdot (5)/(2)+4`

Simplify, we get,

`y_v=-(9)/(4)`

Thus, The vertex of parabola is
`\left((5)/(2),\:-(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)=(-b\pm √(b^2-4ac))/(2a)`

we have a = 1 , b = -5 and c = 4

`x_(1,\:2)=(-\left(-5\right)\pm √(\left(-5\right)^2-4\cdot \:1\cdot \:4))/(2\cdot \:1)`

Simplify, we have,

`x_(1,2)=(5\pm√(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.