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
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(20849) (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