Answer:
y = 6x
Step-by-step explanation:
Given:
[tex]\displaystyle \large{g(x)=4x^2+6x}[/tex]
To find:
- Tangent line equation at x = 0
First, derive the equation using power rules. Here are some power rules formula:
Power Rules
[tex]\displaystyle \large{f(x) = x^n \to f'(x)=nx^{n-1}}\\\\\displaystyle \large{f(x)=kx^n \to f'(x)=knx^{n-1} \quad \tt{(k \ \ is \ \ constant.)}}\\\\\displaystyle \large{f(x)=k \to f'(x)=0 \quad \tt{(k \ \ is \ \ constant.)}[/tex]
Apply power rules:
[tex]\displaystyle \large{g'(x)=4(2)x^{2-1}+6(1)x^{1-1}}\\\\\displaystyle \large{g'(x)=8x+6}[/tex]
Derivative Definition
- Derivative simply means slope or gradient at any points (x,y) and it is also rate of changes.
Since we want to find the slope at x = 0 (so that the line will be tangent to this point) then substitute x = 0 in g’(x):
[tex]\displaystyle \large{g'(0)=8(0)+6}\\\\\displaystyle \large{g'(0)=6}[/tex]
Now we have slope = 6 at x = 0. Next, find y-value at x = 0, simply substitute x = 0 in g(x) to find y-value:
[tex]\displaystyle \large{g(0)=4(0)^2+6(0)}\\\\\displaystyle \large{g(0)=0}[/tex]
So we have point (0,0) which is origin point. Before we head to next step, let’s review on what we have:
- Slope at x = 0 is 6
- Point (0,0)
Next, we use point-slope form to create the equation and convert to slope-intercept form:
Point-Slope
[tex]\displaystyle \large{y-y_1=m(x-x_1)}[/tex]
Determine:
- [tex]\displaystyle \large{m=6}[/tex]
- [tex]\displaystyle \large{(x_1,y_1)=(0,0)}[/tex]
Therefore:
[tex]\displaystyle \large{y-0=6(x-0)}\\\\\displaystyle \large{y=6x}[/tex]
Therefore, the equation of tangent line to the parabola at x = 0 is y = 6x
