How to find the indicated quantities for f(x) = 3x^2? Feb 23, 2017

A) Slope = $6 + 3 h$
B) Slope = 6
C) $y = 6 x - 3$

Explanation:

We have $f \left(x\right) = 3 {x}^{2}$

A. Slope of Secant Line
The slop of the secant line is given by:

$\frac{\Delta y}{\Delta x} = \frac{f \left(1 + h\right) - f \left(1\right)}{\left(1 + h\right) - \left(1\right)}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{3 {\left(1 + h\right)}^{2} - 3}{h}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{3 \left(1 + 2 h + {h}^{2}\right) - 3}{h}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{3 + 6 h + 3 {h}^{2} - 3}{h}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{6 h + 3 {h}^{2}}{h}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = 6 + 3 h$

B. Slope of the graph (tangent) at $\left(1. f \left(1\right)\right)$
If wd take the limit of the slope at the secant line (A) then by the definition of the derivative then in the limit as $h \rightarrow 0$ this will become the slope of the tangent at $x = 1$, So

${\lim}_{h \rightarrow 0} 6 + 3 h = 6$

C. Equation of tangent
Th slope of the tangent is $6$, and it passes through $\left(1 , 3\right)$ so using the point/slope equation of a straight line $y - {y}_{1} = m \left(x - {x}_{1}\right)$ we get:

$\setminus \setminus \setminus \setminus \setminus y - 3 = 6 \left(x - 1\right)$
$\therefore y - 3 = 6 x - 6$
$\therefore \setminus \setminus \setminus \setminus \setminus \setminus \setminus y = 6 x - 3$

Which we can confirm via a graph:
graph{ (y-3x^2)(y-6x+3)=0 [-5, 5, -2, 10]}