Question
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Find and simplify \[ \frac{(a+h)-f(a)}{h},(h \neq 0) \] for the following function. \[ f(x)=4 x^{2}-2 x+7 \] -2

Ask by Warner Sandoval. in the United States
Jan 26,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

The simplified expression is \( \frac{-4a^2 + 3a - 7}{h} + 1 \).

Solution

To simplify the expression \[ \frac{(a+h)-f(a)}{h} \] for the function \[ f(x) = 4x^2 - 2x + 7, \] we will follow these steps: 1. **Calculate \( f(a) \)**: Substitute \( a \) into the function \( f(x) \). 2. **Substitute \( f(a) \) into the expression**: Replace \( f(a) \) in the expression. 3. **Simplify the expression**: Perform algebraic simplifications. ### Step 1: Calculate \( f(a) \) Substituting \( a \) into the function: \[ f(a) = 4a^2 - 2a + 7. \] ### Step 2: Substitute \( f(a) \) into the expression Now, we substitute \( f(a) \) into the expression: \[ \frac{(a+h) - f(a)}{h} = \frac{(a+h) - (4a^2 - 2a + 7)}{h}. \] ### Step 3: Simplify the expression Now, we simplify the numerator: \[ a + h - (4a^2 - 2a + 7) = a + h - 4a^2 + 2a - 7 = -4a^2 + 3a + h - 7. \] Thus, the expression becomes: \[ \frac{-4a^2 + 3a + h - 7}{h}. \] Now, we can separate the terms in the numerator: \[ \frac{-4a^2 + 3a - 7}{h} + \frac{h}{h} = \frac{-4a^2 + 3a - 7}{h} + 1. \] So, the final simplified expression is: \[ \frac{-4a^2 + 3a - 7}{h} + 1. \] This is the simplified form of the given expression.

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Mind Expander

To simplify the expression \(\frac{(a+h) - f(a)}{h}\), we first need to calculate \(f(a)\) for the function \(f(x) = 4x^2 - 2x + 7\). Calculating \(f(a)\): \[ f(a) = 4a^2 - 2a + 7 \] Now, substituting this into the given expression: \[ \frac{(a+h) - f(a)}{h} = \frac{(a+h) - (4a^2 - 2a + 7)}{h} \] Simplifying the numerator: \[ (a + h) - (4a^2 - 2a + 7) = a + h - 4a^2 + 2a - 7 = -4a^2 + 3a + h - 7 \] Now substituting back into the fraction: \[ \frac{-4a^2 + 3a + h - 7}{h} \] This can be separated into two terms: \[ \frac{-4a^2 + 3a - 7}{h} + 1 \] Thus, we rewrite the expression and simplify: \[ \frac{-4a^2 + 3a - 7}{h} + 1 \] Simplifying the final result gives us: \[ \frac{-4a^2 + 3a - 7 + h}{h} \] So, the simplified expression is: \[ \frac{-4a^2 + 3a - 7 + h}{h} \]

Related Questions

Multiple Choice Identify the choice that best completes the statement or answers the question. Find any points of discontinuity for the rational function. 1. \( y=\frac{(x-7)(x+2)(x-9)}{(x-5)(x-2)} \) a. \( x=-5, x=-2 \) b. \( x=5, x=2 \) c. \( x=-7, x=2, x=-9 \) d. \( x=7, x=-2, x=9 \) 2. \( y=\frac{(x+7)(x+4)(x+2)}{(x+5)(x-3)} \) a. \( x=-5, x=3 \) b. \( x=7, x=4, x=2 \) c. \( x=-7, x=-4, x=-2 \) d. \( x=5, x=-3 \) 3. \( y=\frac{x+4}{x^{2}+8 x+15} \) a. \( x=-5, x=-3 \) b. \( x=-4 \) c. \( x=-5, x=3 \) d. \( x=5, x=3 \) 4. \( y=\frac{x-3}{x^{2}+3 x-10} \) a. \( x=-5, x=2 \) b. \( x=5, x=-2 \) c. \( x=3 \) d. \( x \) \( =-5, x=-2 \) 6. What are the points of discontinuity? Are they all removable? \[ y=\frac{(x-4)}{x^{2}-13 x+36} \] a. \( x=-9, x=-4, x=8 \); yes b. \( x=1, x=8, x= \) -8; no c. \( x=9, x=4 \); no d. \( x=-9, x=-4 \); no 7. Describe the vertical asymptote(s) and hole(s) for the graph of \( y=\frac{(x-2)(x-5)}{(x-5)(x+2)} \). a. asymptote: \( x=2 \) and hole: \( x=-5 \) b. asymptotes: \( x=-2 \) and hole: \( x=-5 \) c. asymptote: \( x=-2 \) and hole: \( x=5 \) d. asymptote: \( x=-2 \) and hole: \( x=-2 \) a. \( x=-3, x=-8 \); no b. \( x=5, x=-7, x=1 \); no c. \( x=-5, x=7, x=-1 \); yes d. \( x=3, x=8 \); yes 8. Describe the vertical asymptote(s) and hole(s) for the graph of \( y=\frac{x+1}{x^{2}+6 x+5} \). a. asymptote: \( x=-1 \) and hole: \( x=-1 \) b. asymptote: \( x=-1 \) and hole: \( x=-1 \) c. asymptotes: \( x=-1,-1 \) and hole: \( x=-1 \) d. asymptote: -5 and hole: \( x=-1 \), 9. Find the horizontal asymptote of the graph of \( y=\frac{7 x^{6}+7 x+3}{9 x^{5}+7 x+3} \). a. \( y=0 \) b. \( y=\frac{7}{9} \) c. no horizontal asymptote d. \( y=\frac{6}{5} \)
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