Let's go through the questions one by one.

Question 1:

For $f(x) = 4x^2 - 3x + 1$, evaluate:

1. $f(a)$
2. $f(a + h)$
3. $f(a + h) - f(a)$

• $f(a)$: Substitute $x = a$ into the function: $f(a) = 4a^2 - 3a + 1$

• $f(a + h)$: Substitute $x = a + h$: $f(a + h) = 4(a + h)^2 - 3(a + h) + 1$ Expand the terms:

  = 4a^2 + 8ah + 4h^2 - 3a - 3h + 1$$

• $f(a + h) - f(a)$: Subtract $f(a)$ from $f(a + h)$: $f(a + h) - f(a) = \left(4a^2 + 8ah + 4h^2 - 3a - 3h + 1\right) - \left(4a^2 - 3a + 1\right)$ Simplifying: $= 8ah + 4h^2 - 3h$

Question 2:

For $f(x) = \frac{1}{x}$, evaluate:

1. $f(a)$
2. $f(a + h)$
3. $f(a + h) - f(a)$

• $f(a)$: $f(a) = \frac{1}{a}$

• $f(a + h)$: $f(a + h) = \frac{1}{a + h}$

• $f(a + h) - f(a)$: $f(a + h) - f(a) = \frac{1}{a + h} - \frac{1}{a}$ Use a common denominator to simplify: $f(a + h) - f(a) = \frac{a - (a + h)}{a(a + h)} = \frac{-h}{a(a + h)}$

Question 3:

For $f(x) = \sqrt{3 - 2x}$, find the:

1. x-intercept: Set $f(x) = 0$, which gives: $\sqrt{3 - 2x} = 0 \implies 3 - 2x = 0 \implies x = \frac{3}{2}$ So, the x-intercept is $\left(\frac{3}{2}, 0\right)$.

2. y-intercept: Set $x = 0$, and evaluate $f(0)$: $f(0) = \sqrt{3 - 2(0)} = \sqrt{3}$ So, the y-intercept is $(0, \sqrt{3})$.

Question 4:

Find the domain and range for $y = \sqrt{8x - 1} - 1$.

• Domain: The expression inside the square root must be non-negative: $8x - 1 \geq 0 \implies x \geq \frac{1}{8}$ So, the domain is $\left[\frac{1}{8}, \infty \right)$.

• Range: The function $\sqrt{8x - 1}$ has a minimum value of 0, so the range of $y = \sqrt{8x - 1} - 1$ is: $y \geq -1$ Therefore, the range is $[-1, \infty)$.

Questions 5 to 8:

Would you like me to continue with the rest of the questions or clarify anything from the above steps?

Related Questions:

1. How do we evaluate functions at $f(a+h)$ for more complex functions?
2. How do we find the domain and range for functions involving fractions?
3. What are the conditions for a function to have an x-intercept?
4. Can you explain why square roots impact the domain of functions?
5. How would the x-intercept calculation differ if the function were $\sqrt{x^2 - 4x}$?

Tip:

Always check the domain of a function before performing further calculations, especially when dealing with square roots and fractions!