Part (a)
Plug in y = 0 and solve for x. Use the zero product property
y = x(x+3)(x-2)
0 = x(x+3)(x-2)
x(x+3)(x-2) = 0
x = 0 or x+3 = 0 or x-2 = 0 .... zero product property
x = 0 or x = -3 or x = 2
The three roots or zeros are x = 0 or x = -3 or x = 2
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Part (b)
The roots of the graph are: x = -2, x = 1, x = 3. Each root is where the graph crosses or touches the horizontal x axis.
Note how x = 0 is found in part (a), but not found here. This is one example where the graphs don't match. Another would be x = -3 is in part (a), but not here.
So that's why the graph does <u>not</u> match with the function in part (a)
The function notation of the following are:
f(1) = 5
f(0) = 1
f(-3) = -11
f(8) = 33
Given:
x is the input.
the function is 4x+1
f(x) is the output.
we are asked to determine the values of the following :
a. f(1)
x = 1
so f(x) = 4x+1
f(1) = 4(1)+1
f(1) = 4+1
f(1)=5
b. f(0)
x = 0
so f(x) = 4x+1
f(0) = 4(0)+1
f(0) = 0+1
f(0)=1
c. f(-3)
x = -3
so f(x) = 4x+1
f(-3) = 4(-3)+1
f(1) = -12+1
f(1)= -11
d. f(8)
x = 8
so f(x) = 4x+1
f(8) = 4(8)+1
f(8) = 32+1
f(1)=33
Hence we get the required values.
Learn more about Functions here:
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Answer:
63%
Step-by-step explanation:
<em>From the question, we aim to find the percent increase in the tuition</em>
Given data
initial cost= $99 per credit hour
Final cost= $268 per credit hour
% increase= (Final - initial )/initial *100
substitute
% increase= (268- 99 )/268 *100
% increase= 169 /268 *100
% increase= 0.630*100
% increase= 63%
Hence the increase in the tuition from 1990 to 2003 is 63%
Answer:
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Step-by-step explanation: