For the function \( f(x)=5 x+x^{2} \), find each of the following. a) \( f(t-2) \) b) \( f(p+2) \) c) \( f(a+h)-f(a) \) d) \( f(t-1)+c \) e) \( f(a)+7 \) a) \( f(t-2)=t^{2}+t-6 \) (Simplify your answer. Do not factor.) b) \( f(p+2)=p^{2}+9 p+14 \) (Simplify your answer. Do not factor.) c) \( f(a+h)-f(a)=\square \) (Simplify your answer. Do not factor.)

To work with the function \( f(x) = 5x + x^2 \) and find each of the requested expressions, let’s go through each one step by step. a) **Finding \( f(t-2) \):** We substitute \( t-2 \) into the function: \[ f(t-2) = 5(t-2) + (t-2)^2 \] Calculating each part: \[ = 5t - 10 + (t^2 - 4t + 4) \] Combining terms: \[ = t^2 + t - 6 \] So, \( f(t-2) = t^2 + t - 6 \). b) **Finding \( f(p+2) \):** We substitute \( p+2 \) into the function: \[ f(p+2) = 5(p+2) + (p+2)^2 \] Calculating: \[ = 5p + 10 + (p^2 + 4p + 4) \] Combining: \[ = p^2 + 9p + 14 \] So, \( f(p+2) = p^2 + 9p + 14 \). c) **Finding \( f(a+h) - f(a) \):** First, we calculate \( f(a+h) \): \[ f(a+h) = 5(a+h) + (a+h)^2 \] Calculating: \[ = 5a + 5h + (a^2 + 2ah + h^2) \] This simplifies to: \[ = a^2 + 5a + 5h + 2ah + h^2 \] Now, we also need \( f(a) \): \[ f(a) = 5a + a^2 \] Subtracting \( f(a) \) from \( f(a+h) \): \[ f(a+h) - f(a) = (a^2 + 5a + 5h + 2ah + h^2) - (5a + a^2) \] The \( 5a \) and \( a^2 \) terms cancel out: \[ = 5h + 2ah + h^2 \] So, \( f(a+h) - f(a) = 5h + 2ah + h^2 \). d) **Finding \( f(t-1) + c \):** We substitute \( t-1 \): \[ f(t-1) = 5(t-1) + (t-1)^2 \] Calculating: \[ = 5t - 5 + (t^2 - 2t + 1) = t^2 + 3t - 4 \] Adding \( c \): \[ f(t-1) + c = t^2 + 3t - 4 + c \] So, \( f(t-1) + c = t^2 + 3t - 4 + c \). e) **Finding \( f(a) + 7 \):** Since we already found \( f(a) = 5a + a^2 \): \[ f(a) + 7 = 5a + a^2 + 7 \] So, \( f(a) + 7 = a^2 + 5a + 7 \). Putting it all together: - a) \( f(t-2) = t^2 + t - 6 \) - b) \( f(p+2) = p^2 + 9p + 14 \) - c) \( f(a+h) - f(a) = 5h + 2ah + h^2 \) - d) \( f(t-1) + c = t^2 + 3t - 4 + c \) - e) \( f(a) + 7 = a^2 + 5a + 7 \)