To find the equation of the tangent line to the graph of $f(x) = 4x^2 - 3$ at the given point $(0, -3)$, we will use the limit definition of the derivative:

$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Step 1: Compute $f'(x)$ using the limit definition

We start with:

$f(x) = 4x^2 - 3$

Using the definition of the derivative:

$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Substituting $f(x)$:

$f'(x) = \lim_{h \to 0} \frac{4(x+h)^2 - 3 - (4x^2 - 3)}{h}$

Expanding $(x+h)^2$:

$(x+h)^2 = x^2 + 2xh + h^2$

$4(x+h)^2 = 4x^2 + 8xh + 4h^2$

$f(x+h) = 4x^2 + 8xh + 4h^2 - 3$

Now, compute the difference:

$f(x+h) - f(x) = (4x^2 + 8xh + 4h^2 - 3) - (4x^2 - 3)$

$= 4x^2 + 8xh + 4h^2 - 3 - 4x^2 + 3$

$= 8xh + 4h^2$

Now, divide by $h$:

$\frac{8xh + 4h^2}{h} = 8x + 4h$

Taking the limit as $h \to 0$:

$f'(x) = 8x$

Step 2: Find the slope of the tangent line at $x = 0$

$f'(0) = 8(0) = 0$

Since the slope of the tangent line is 0, the line is horizontal.

Step 3: Find the equation of the tangent line

Using the point-slope form of a line:

$y - y_1 = m(x - x_1)$

$y - (-3) = 0(x - 0)$

$y + 3 = 0$

$y = -3$

So, the equation of the tangent line is:

$y = -3$

Step 4: Verify with a Graphing Utility

You can graph $f(x) = 4x^2 - 3$ and the horizontal line $y = -3$ using a graphing calculator or a tool like Desmos or GeoGebra.

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Related Questions

1. How do you interpret the derivative as a rate of change in this example?
2. What does a horizontal tangent line indicate about the function at that point?
3. How would the equation change if the function were $f(x) = 4x^2 + 3$?
4. Can you use implicit differentiation for this problem? Why or why not?
5. What is the second derivative of $f(x)$, and what does it tell us?

Math Tip:

A horizontal tangent line means that the function has a local minimum or maximum at that point. In this case, $(0, -3)$ is the lowest point on the graph, meaning it's a local minimum.