The instantaneous rate of change of the function f(x) is h > 0 (f(x + h) - f(x))/h substituting f(x) 6x^2 + 9x into this formula getting rid of the parentheses, we have the following:
Lim h > 0 f(x) is h > 0 (f(x + h) - f(x))/h = lim h > 0 (6(x+h)^2 + 9(h+x) - (6x^2 + 9x)/h = lim > 0 (6x^2 + 12(hx) + 6h^2 + 9x + 9h -6x^2 - 9x)/h

Simplifying the numerator, factoring out the common factor h, and then canceling the factor h with the denominator h, we have the following: lim > 0 (6x^2 + 12xh + 6h^2 + 9x + 9h - 6x^2 - 9x)/h = lim h > 0 (12(blank) +6h^2 + blank)/h

Math = lim > 0 blank

Give me every blank

Let's break down this problem and solve for each "blank."

### Step 1: Function definition and difference quotient
You are given the function:
\[
f(x) = 6x^2 + 9x
\]
The difference quotient formula for the instantaneous rate of change is:
\[
\frac{f(x+h) - f(x)}{h}
\]
Substitute \( f(x) = 6x^2 + 9x \) into this formula:
\[
\frac{(6(x+h)^2 + 9(x+h)) - (6x^2 + 9x)}{h}
\]

### Step 2: Expand \( f(x+h) \)
Now, let's expand \( f(x+h) = 6(x+h)^2 + 9(x+h) \):

1. \( 6(x+h)^2 = 6(x^2 + 2xh + h^2) = 6x^2 + 12xh + 6h^2 \)
2. \( 9(x+h) = 9x + 9h \)

So, \( f(x+h) = 6x^2 + 12xh + 6h^2 + 9x + 9h \).

### Step 3: Simplify the difference quotient
Now subtract \( f(x) = 6x^2 + 9x \) from \( f(x+h) \):
\[
\frac{(6x^2 + 12xh + 6h^2 + 9x + 9h) - (6x^2 + 9x)}{h}
\]

Simplifying the numerator:
\[
6x^2 + 12xh + 6h^2 + 9x + 9h - 6x^2 - 9x = 12xh + 6h^2 + 9h
\]

So, the expression becomes:
\[
\frac{12xh + 6h^2 + 9h}{h}
\]

### Step 4: Factor and cancel
Factor \( h \) from the numerator:
\[
\frac{h(12x + 6h + 9)}{h}
\]
Cancel the \( h \) terms:
\[
12x + 6h + 9
\]

### Step 5: Take the limit as \( h \to 0 \)
Now, take the limit as \( h \to 0 \):
\[
\lim_{h \to 0} (12x + 6h + 9) = 12x + 9
\]

### Final Answer for Blanks:
- First blank: \( x \) (from \( 12x \))
- Second blank: \( h \) (from \( 6h \))

Thus, the final expression is:
\[
\lim_{h \to 0} (12x + 6h + 9) = 12x + 9
\]

---

Let me know if you'd like further details or clarifications. Here are five related questions you might find useful:

1. What is the general formula for the derivative using limits?
2. How does the difference quotient relate to the derivative of a function?
3. What is the interpretation of the derivative in terms of rates of change?
4. What happens when you differentiate a constant function?
5. How does the derivative of a polynomial behave in general?

**Tip:** Always check your algebraic expansions carefully when calculating limits to avoid small errors!

The instantaneous rate of change of the function f(x) is h > 0 (f(x + h) - f(x))/h substituting f(x) 6x^2 + 9x into this formula getting rid of the parentheses, we have the following: Lim h > 0 f(x) is h > 0 (f(x + h) - f(x))/h = lim h > 0 (6(x+h)^2 + 9(h+x) - (6x^2 + 9x)/h = lim > 0 (6x^2 + 12(hx) + 6h^2 + 9x + 9h -6x^2 - 9x)/h. Simplifying the numerator, factoring out the common factor h, and then canceling the factor h with the denominator h, we have the following: lim > 0 (6x^2 + 12xh + 6h^2 + 9x + 9h - 6x^2 - 9x)/h = lim h > 0 (12(blank) +6h^2 + blank)/h. Math = lim > 0 blank. Give me every blank.

This detailed solution explains how to calculate the instantaneous rate of change of the quadratic function f(x) = 6x^2 + 9x using the difference quotient and limits. Follow a step-by-step breakdown of the limit process, including expansion, simplification, and factoring techniques. Suitable for students in Grades 11-12 or introductory college calculus.

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