You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
Intermediate Algebra
- Thread starter jamjam
- Start date
jamjam said:Find the x- & y- intercepts.If no x- intercepts exist,state so.
f(x)=-x^2+11x-30
Hi jamjam,
In \(\displaystyle f(x)=ax^2+bx+c\), the y-intercept = c.
The x-intercepts are found by solving\(\displaystyle -x^2+11x-30=0\)
jamjam said:Find the x- & y- intercepts.If no x- intercepts exist,state so.
f(x)=-x^2+11x-30
IN GENERAL, a y-intercept is the value of the function when x = 0, because when a point is on the y-axis, the x-coordinate of the point must be 0.
So, if you've got a function of x and you're looking for the y-intercept, let x = 0, and solve for y.
IN GENERAL, an x-intercept is a point on the x-axis...and every point on the x-axis has a y-coordinate of 0. If you've got a function of x, and you're looking for the x-intercept(s), let f(x) = 0, and solve for the x value(s) which make this true.