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November 28, 2014

November 28, 2014

Posted by **-Untamed-** on Thursday, November 24, 2011 at 9:14pm.

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)

- Math -
**Steve**, Friday, November 25, 2011 at 12:05amIt'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

= 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.

_______________________

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)

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