SOLUTION: Thank you Stan for answering this the first time, however, I did fail to put in that we have to find the x and y intercepts. f(x)=x^2-9x+5 I used the -b/2a and 4ac-b^2/4a formula

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Question 176602: Thank you Stan for answering this the first time, however, I did fail to put in that we have to find the x and y intercepts.
f(x)=x^2-9x+5
I used the -b/2a and 4ac-b^2/4a formulas which makes the problems 9/2(1)and 4(1)(5)-(-9)^2/4(1).
I came up with x=4.5 and y=25.25 Is this right? Thank you so much.
Found 2 solutions by MathLover1, Mathtut:
Answer by MathLover1(20855) (Show Source):
You can put this solution on YOUR website!
the intercept- where

Solved by pluggable solver: Quadratic Formula

Let's use the quadratic formula to solve for x:

Starting with the general quadratic

the general solution using the quadratic equation is:

So lets solve ( notice , , and )

Plug in a=1, b=-9, and c=5

Negate -9 to get 9

Square -9 to get 81 (note: remember when you square -9, you must square the negative as well. This is because .)

Multiply to get

Combine like terms in the radicand (everything under the square root)

Simplify the square root (note: If you need help with simplifying the square root, check out this solver)

Multiply 2 and 1 to get 2

So now the expression breaks down into two parts

or

Now break up the fraction

or

Simplify

or

So the solutions are:

or

the intercept- where

Solved by pluggable solver: PLOT a graph of a function

Plotting a graph:

This solver uses formula rendering system and here's the actual formula that was plotted:

graph( 600, 400, -15, 15, -15, 15, x^2-9x+5 )

Answer by Mathtut(3670) (Show Source):
You can put this solution on YOUR website!
Find the x-intercepts:
Notice that the x-intercepts of any graph are points on the x-axis and therefore have y-coordinate 0. We can find these points by plugging 0 in for y and solving the resulting quadratic equation (0 = ax2 + bx + c). If the equation factors we can find the points easily, but we may have to use the quadratic formula in some cases. If the solutions are imaginary, that means that the parabola has no x-intercepts (is strictly above or below the x-axis and never crosses it). If the solutions are real, but irrational (radicals) then we need to approximate their values and plot them.
Find the y-intercept:
The y-intercept of any graph is a point on the y-axis and therefore has x-coordinate 0. We can use this fact to find the y-intercepts by simply plugging 0 for x in the original equation and simplifying. So the y-intercept of any parabola is always at (0,c).
f(x)=x^2-9x+5
:
setting x to zero
:

:
so y=c or y=5
:
x=8.49 and x=.59

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=61 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 8.40512483795333, 0.594875162046673. Here's your graph: