Replace the [latex]x[/latex] in the function with each specified value.

1. Because the input value is a number, 2, we can use algebra to simplify.

   [latex]\begin{align}f\left(2\right)&={2}^{2}+3\left(2\right)-4 \\ &=4+6 - 4 \\ &=6\hfill \end{align}[/latex]

2. In this case, the input value is a letter so we cannot simplify the answer any further.

   [latex]f\left(a\right)={a}^{2}+3a - 4[/latex]

3. With an input value of [latex]a+h[/latex], we must use the distributive property.

   [latex]\begin{align}f\left(a+h\right)&={\left(a+h\right)}^{2}+3\left(a+h\right)-4 \\[2mm] &={a}^{2}+2ah+{h}^{2}+3a+3h - 4 \end{align}[/latex]

4. In this case, we apply the input values to the function more than once, and then perform algebraic operations on the result. We already found that:

   [latex]f\left(a+h\right)={a}^{2}+2ah+{h}^{2}+3a+3h - 4[/latex]

   and we know that:

   [latex]f\left(a\right)={a}^{2}+3a - 4[/latex]

   Now we combine the results and simplify.

   [latex]\begin{align}\dfrac{f\left(a+h\right)-f\left(a\right)}{h}&=\dfrac{\left({a}^{2}+2ah+{h}^{2}+3a+3h - 4\right)-\left({a}^{2}+3a - 4\right)}{h} \\[2mm] &=\dfrac{2ah+{h}^{2}+3h}{h}\\[2mm] &=\frac{h\left(2a+h+3\right)}{h}&&\text{Factor out }h. \\[2mm] &=2a+h+3&&\text{Simplify}.\end{align}[/latex]

Substitute [latex]-3[/latex] in for [latex]x[/latex] in the function.

[latex]p(−3)=2(−3)^{2}+5[/latex]

Simplify the expression on the right side of the equation.

[latex]\begin{array}{lll}p(−3)=2(9)+5\\p(−3)=18+5\\p(−3)=23\end{array}[/latex]

[latex]\begin{align}&h\left(p\right)=3\\ &{p}^{2}+2p=3 &&\text{Substitute the original function }h\left(p\right)={p}^{2}+2p. \\ &{p}^{2}+2p - 3=0 &&\text{Subtract 3 from each side}. \\ &\left(p+3\text{)(}p - 1\right)=0 &&\text{Factor}. \end{align}[/latex]

If [latex]\left(p+3\right)\left(p - 1\right)=0[/latex], either [latex]\left(p+3\right)=0[/latex] or [latex]\left(p - 1\right)=0[/latex] (or both of them equal [latex]0[/latex]). We will set each factor equal to [latex]0[/latex] and solve for [latex]p[/latex] in each case.

[latex]\begin{align}&p+3=0, &&p=-3 \\ &p - 1=0, &&p=1\hfill \end{align}[/latex]

This gives us two solutions. The output [latex]h\left(p\right)=3[/latex] when the input is either [latex]p=1[/latex] or [latex]p=-3[/latex].

 

We can also verify by graphing, as seen above. The graph verifies that [latex]h\left(1\right)=h\left(-3\right)=3[/latex] and [latex]h\left(4\right)=24[/latex].