Given the function:
![f(x)=x^2-3x](https://img.qammunity.org/2023/formulas/mathematics/college/3v7mubvgnhbkqsouvb15u74m1or9904myk.png)
Let's solve for f(5).
To solve the function f when we have the indicated value f(5), substitute 5 for x in the function and solve.
Thus, we have:
![\begin{gathered} f(5)=5^2-3(5) \\ \\ f(5)=25-15 \\ \\ f(5)=10 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/slorbjpl37f64jykxucbcf0jn3eo000iwn.png)
Let's solve for f(x) = 4.
To solve when f(x) = 4, substitute 4 for f(x) amd solve for the values of x.
Thus, we have:
![\begin{gathered} 4=x^2-3x \\ \\ x^2-3x=4 \\ \\ \text{Subtract 4 from both sides:} \\ x^2-3x-4=4-4 \\ \\ x^2-3x-4=0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/9yuwgorlha6gx9qwv0j2xoli13ii6rxdfn.png)
Factorize the left side using the AC method:
Find a pair of numbers whose product is 4 and sum is -3.
We have:
-4 and 1
![(x-4)(x+1)=0](https://img.qammunity.org/2023/formulas/mathematics/college/h7wzkeu6fo3nurov5wtdit77iwcdrlgdj1.png)
Set each factor to zero and solve for x:
![\begin{gathered} x-4=0 \\ \text{Add 4 to both sides:} \\ x-4+4=0+4 \\ x=4 \\ \\ x+1=0 \\ \text{Subtract 1 from both sides:} \\ x+1-1=0-1 \\ \\ x=-1 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/39r17r5s3z67frndr7d1rwkj69riu9f16f.png)
Therefore, if f(x) = 4, then x = 4 and -1.
The grap represents a function because it passes the vertical line test.
The graph does not represent a one-to-one function.
ANSWER:
• f(5) = 10
,
• If f(x) = 4 then x = 4 and -1
,
• The graph represents a function.
,
• The graph does not represent a one-to-one function.