To solve a, we need to replace x = 0 in the formula of the function:
[tex]\begin{cases}f(x)=x^2+5 \\ x=0\end{cases}\Rightarrow f(0)=0^2+5=5[/tex]The y value when x = 0 is 5.
b is asking the same as a but in a different way. The y-intercept of a function is when x = 0, we just calculated that. The point of y-intercept is (0, 5)
Finally, to solve c, we need to find the values of x that gives us a value of f(x) = 46:
[tex]f(x)=46\Rightarrow46=x^2+5[/tex]Then solve:
[tex]\begin{gathered} x^2=46-5 \\ x=\pm\sqrt[]{41} \end{gathered}[/tex]Remember that we must that plus-minus the value when we take square root. ± √41 is the answer to c.
