jpalmer11 wrote:For f(x)=x^2-6x+1
I think the reason I can not find the X and Y intercepts is there are no intercepts. Is that correct.
To find the x-intercept(s), plug zero in for y (in this case, for f(x)) and solve. What do you get?
To find the y-intercept, plug zero in for x and solve. What do you get?
Note: You can also look at the graph to confirm
the intercept(s), if any.
jpalmer11 wrote:on the second problem solve for x
x^2+7x-4=0
can you just let me know if I am heading in the right direction with
- (-10) + 2 the square of 10^2 - 164 devided by 2
Your
formatting is unclear, and I have no idea what is meant by the squared "plus" sign. Do you maybe mean the following?
. . . . .(-(-10) +/- sqrt[10^2 - 164]) / 2
...which typesets as:
. . . . .\, \pm\, \sqrt{10^2\, -\, 164}}{2})
If so, how did you get this? I'm not seeing how this relates to
the Quadratic Formula or any other method of solving this quadratic equation (which has a = 1, b = 7, and c = -4)...?