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Question
Math
Posted 7 months ago

Let ff be the function defined by the rule f(x)=x2x5f(x)=x^{2}-x-5.
(a) Find the domain of ff. (Enter your answer using interval notation.)
(b) Compute f(x)f(x) for x=3,2,1,0,12,1,2,3x=-3,-2,-1,0, \frac{1}{2}, 1,2,3.
f(3)=f(2)=f(1)=f(0)=f(12)=f(1)=f(2)=f(3)=
\begin{aligned}
f(-3) & = \\
f(-2) & = \\
f(-1) & = \\
f(0) & = \\
f\left(\frac{1}{2}\right) & = \\
f(1) & = \\
f(2) & = \\
f(3) & =
\end{aligned}

\square
\square
\square
\square
\square
\square
\square
\square
(c) Use the results obtained in parts (a) and (b) to sketch the graph of ff.
(i)
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Answer from Sia
Posted 7 months ago
Solution by Steps
step 1
To find the domain of the function f(x)=x2x5f(x) = x^2 - x - 5, we note that it is a polynomial function. Polynomial functions are defined for all real numbers. Therefore, the domain of ff is all real numbers
step 2
To compute f(x)f(x) for x=3,2,1,0,12,1,2,3x = -3, -2, -1, 0, \frac{1}{2}, 1, 2, 3, we substitute each value of xx into the function f(x)=x2x5f(x) = x^2 - x - 5
step 3
For x=3x = -3: f(3)=(3)2(3)5=9+35=7 f(-3) = (-3)^2 - (-3) - 5 = 9 + 3 - 5 = 7
step 4
For x=2x = -2: f(2)=(2)2(2)5=4+25=1 f(-2) = (-2)^2 - (-2) - 5 = 4 + 2 - 5 = 1
step 5
For x=1x = -1: f(1)=(1)2(1)5=1+15=3 f(-1) = (-1)^2 - (-1) - 5 = 1 + 1 - 5 = -3
step 6
For x=0x = 0: f(0)=0205=5 f(0) = 0^2 - 0 - 5 = -5
step 7
For x=12x = \frac{1}{2}: f(12)=(12)2125=14125=214 f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - \frac{1}{2} - 5 = \frac{1}{4} - \frac{1}{2} - 5 = -\frac{21}{4}
step 8
For x=1x = 1: f(1)=1215=115=5 f(1) = 1^2 - 1 - 5 = 1 - 1 - 5 = -5
step 9
For x=2x = 2: f(2)=2225=425=3 f(2) = 2^2 - 2 - 5 = 4 - 2 - 5 = -3
step 10
For x=3x = 3: f(3)=3235=935=1 f(3) = 3^2 - 3 - 5 = 9 - 3 - 5 = 1
step 11
Using the results from steps 3 to 10, we can sketch the graph of f(x)=x2x5f(x) = x^2 - x - 5. The points to plot are: (3,7)(-3, 7), (2,1)(-2, 1), (1,3)(-1, -3), (0,5)(0, -5), (12,214)\left(\frac{1}{2}, -\frac{21}{4}\right), (1,5)(1, -5), (2,3)(2, -3), (3,1)(3, 1)
Answer
The domain of ff is all real numbers. The computed values are f(3)=7f(-3) = 7, f(2)=1f(-2) = 1, f(1)=3f(-1) = -3, f(0)=5f(0) = -5, f(12)=214f\left(\frac{1}{2}\right) = -\frac{21}{4}, f(1)=5f(1) = -5, f(2)=3f(2) = -3, f(3)=1f(3) = 1. The graph can be sketched using these points.
Key Concept
Domain and range of polynomial functions
Explanation
Polynomial functions are defined for all real numbers, and their graphs can be sketched by plotting points obtained from evaluating the function at specific values of xx.

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