It's not so different. It just has a fraction. You can always use y instead of f(x)
y = x/3 + 5
To solve, set y=0 and solve for x
0 = x/3 + 5
x/3 = -5
x = -3/5
The y-intercept is always easy. Just plug in x=0 and evaluate y
y = 5x^2 - 35x
y(0) = 0 so the y-intercept is (0,0)
5x^2 - 35x
so, y=0 when
5x(x-7) = 0
x=0 or x=7
The x-intercepts are (0,0) and (7,0)
Note that (0,0) is both an x-intercept and a y-intercept.
y = 3x(x^2 - 49)
Think back to your factoring exercises, and recall the difference of two squares:
(a+b)(a-b) = a^2 - b^2
y = 3x(x+7)(x-7)
So, the y-intercept is (0,0)
x-intercepts are (0,0) (7,0) and (-7,0)
advanced functions/precalculus - 1. The function f(x) = (2x + 3)^7 is the ...
math - i dont understand how this is supposed to be layed out and how to do it ...
Algebra 2 - One of the the zeros of the functions F(x)= x^4+2x^3-13x^2-38x-24 is...
functions - a) find the zeros of t(x) =2x^2 + ix + 3. b) The zeros are not ...
algebra - p(x)=x^3+2x^2-3x+20 one of this functions zeros is -4 When using ...
Maths - I have a test tomorrow and I really don't get this, please help me. "Use...
Math - Learning about quadratic functions and equations, and I am struggling. It...
college algebra - 1. solve the following logarithmic equation log_8(x+8)+log_8(x...
math - find the real zeros of the given functions, use the real zeros to factor ...
7th grade, math - Can someone explain the steps to solve this problem? If a ...