Mathematics Algebra
Find the x-intercepts and y-intercepts for f(x)
f(x)=x^2-4x + 1
I have been trying to figure this one out for hours and I find no solution but there has to be a way to figure this out and I can't. Please help!! Thank you
You can ask a new question or answer this question .
Similar Questions
Top answer:
correct, at the x-intercept, the value of y = 0 other ways: you could graph it and verify those
Read more.
Top answer:
correct, except in your last part ... 0 = x(x + 3) so x = 0 or x = -3 so you have 2 x intercepts,
Read more.
Top answer:
To find the x-intercepts of the equation y = x^2 + 5x + 2, you are correct in stating that the
Read more.
Did I do this right? Find the vertex, the x intercepts and y intercepts. f(x)=(x-2)^2-4 Vertex: 2,4 X intercepts: 0,0 4,0 Y
Top answer:
Isn't the vertex at 0,-4? y intercepts are when x is zero y=-2^2-4 or 0 x intercepts are when y is
Read more.
Top answer:
let y = 2x^2+3x-1 for y-intercept , let x = 0 That one is real easy for x-intercept let y = 0
Read more.
Top answer:
The graphs have different x-intercepts, but the same y-intercept.
Read more.
Top answer:
For the equation 6y = -x + 2, to find the x-intercept, we need to put y = 0 and solve for x. To do
Read more.
Top answer:
To find the y-intercept, you can set x = 0 in the equation y = 0.5x^2 + 2x + 1: y = 0.5(0)^2 + 2(0)
Read more.
R(x)=x^2+x-6 ________ x-3 1. find the domain and y-intercepts 2. find the x-intercepts 3. find the real solution of the equation
Top answer:
R(x) = (x^2 + x - 6)/(x-3) = (x+3)(x-2)/(x-3) 1. domain: any real number, x ≠ 3 y-intercept, let x
Read more.