Find and simplify each of the following for \( f(x)=6 x^{2}-8 x+9 \) (A) \( f(x+h) \) (B) \( f(x+h)-f(x) \) (C) \( \frac{f(x+h)-f(x)}{h} \) (A) \( f(x+h)=\square \) (Do not factor.) (B) \( f(x+h)-f(x)=\square \) (Do not factor.) (C) \( \frac{f(x+h)-f(x)}{h}=\square \) (Do not factor.)

To find \( f(x+h) \), \( f(x+h)-f(x) \), and \( \frac{f(x+h)-f(x)}{h} \) for the function \( f(x) = 6x^2 - 8x + 9 \), let's work step by step. ### (A) \( f(x+h) \) Substituting \( x+h \) into the function, we get: \[ f(x+h) = 6(x+h)^2 - 8(x+h) + 9 \] Calculating \( (x+h)^2 = x^2 + 2xh + h^2 \), we can expand further: \[ f(x+h) = 6(x^2 + 2xh + h^2) - 8x - 8h + 9 \] Thus: \[ f(x+h) = 6x^2 + 12xh + 6h^2 - 8x - 8h + 9 \] Combining like terms yields: \[ f(x+h) = 6x^2 - 8x + 9 + 12xh + 6h^2 - 8h \] So: \[ f(x+h) = 6x^2 - 8x + 9 + 12xh + 6h^2 - 8h \] ### (B) \( f(x+h) - f(x) \) Now we find: \[ f(x+h) - f(x) = (6x^2 - 8x + 9 + 12xh + 6h^2 - 8h) - (6x^2 - 8x + 9) \] Simplifying gives: \[ f(x+h) - f(x) = 12xh + 6h^2 - 8h \] ### (C) \( \frac{f(x+h)-f(x)}{h} \) Now we take the expression from part B and divide by \( h \): \[ \frac{f(x+h) - f(x)}{h} = \frac{12xh + 6h^2 - 8h}{h} \] This simplifies (with \( h \neq 0 \)) to: \[ \frac{f(x+h) - f(x)}{h} = 12x + 6h - 8 \] Final answers: (A) \( f(x+h) = 6x^2 - 8x + 9 + 12xh + 6h^2 - 8h \) (B) \( f(x+h)-f(x) = 12xh + 6h^2 - 8h \) (C) \( \frac{f(x+h)-f(x)}{h} = 12x + 6h - 8 \)