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)
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
= 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)
To find the zeros of a function, we need to solve for the values of x when f(x) is equal to zero. The zeros are the values of x where the graph of the function intersects the x-axis.
Let's start with the first function f(x) = x/3 +5. To find the zeros, we set f(x) equal to zero and solve for x:
0 = x/3 + 5
To isolate x, we subtract 5 from both sides:
-5 = x/3
Then, we multiply both sides by 3 to get rid of the fraction:
-15 = x
So, the zero of the function f(x) = x/3 + 5 is x = -15.
Moving on to the second function f(x) = 5x^2 - 35x. To find the zeros, we set f(x) equal to zero and solve for x:
0 = 5x^2 - 35x
Factoring out 5x from both terms on the right-hand side:
0 = 5x(x - 7)
Now, we can set each factor equal to zero and solve for x:
5x = 0 --> x = 0
x - 7 = 0 --> x = 7
So, the zeros of the function f(x) = 5x^2 - 35x are x = 0 and x = 7.
Lastly, for the function f(x) = 3x(x^2-49), we factor out the common factor of 3x:
f(x) = 3x(x+7)(x-7)
To find the zeros, we set each factor equal to zero and solve for x:
3x = 0 --> x = 0
x + 7 = 0 --> x = -7
x - 7 = 0 --> x = 7
The zeros of the function f(x) = 3x(x^2-49) are x = 0, x = -7, and x = 7.
To determine the y-intercept, we can substitute x = 0 into the function and calculate f(0). The y-intercept is the point where the graph intersects the y-axis.
For the first function, f(x) = x/3 + 5, we substitute x = 0:
f(0) = 0/3 + 5 = 5
So, the y-intercept for f(x) = x/3 + 5 is (0, 5).
For the second function, f(x) = 5x^2 - 35x, when we substitute x = 0:
f(0) = 5(0)^2 - 35(0) = 0
So, the y-intercept for f(x) = 5x^2 - 35x is (0, 0).
Lastly, for the function f(x) = 3x(x^2-49):
f(0) = 3(0)(0^2-49) = 0
So, the y-intercept for f(x) = 3x(x^2-49) is (0, 0).