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
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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
= 5x(x-7)
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.
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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)
Find the Zeros of the following functions
f(x) = x/3 +5
^ That one is layed out so differently I don't understand how to solve.
In each of the following, determine the zeros of the function and determine the y-intercept of the graph of the function.
f(x) = 5x^2 - 35x
f(x) = 3x(x^2-49)
1 answer
1 answer