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TiliK225 [7]
2 years ago
7

Find the equation of the tangent line to g(x) = 4x² + 6x at x = 0 Simplify any fractions

Mathematics
2 answers:
jolli1 [7]2 years ago
8 0

Answer:

y = 6x

Step-by-step explanation:

Given:

\displaystyle \large{g(x)=4x^2+6x}

To find:

  • Tangent line equation at x = 0

First, derive the equation using power rules. Here are some power rules formula:

<u>Power Rules</u>

<u />\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.)}

Apply power rules:

\displaystyle \large{g'(x)=4(2)x^{2-1}+6(1)x^{1-1}}\\\\\displaystyle \large{g'(x)=8x+6}

<u>Derivative Definition</u>

  • 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):

\displaystyle \large{g'(0)=8(0)+6}\\\\\displaystyle \large{g'(0)=6}

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:

\displaystyle \large{g(0)=4(0)^2+6(0)}\\\\\displaystyle \large{g(0)=0}

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:

<u>Point-Slope</u>

<u />\displaystyle \large{y-y_1=m(x-x_1)}

Determine:

  • \displaystyle \large{m=6}
  • \displaystyle \large{(x_1,y_1)=(0,0)}

Therefore:

\displaystyle \large{y-0=6(x-0)}\\\\\displaystyle \large{y=6x}

Therefore, the equation of tangent line to the parabola at x = 0 is y = 6x

Marina CMI [18]2 years ago
6 0

Answer:

y = 6x

To find equation,

1) Find Derivative

2) Find the slope at x

3) Find coordinates

<h3><u>Lets first Derive The Equation</u></h3>

\sf \rightarrow y = 4x^2 + 6x

\sf \rightarrow \dfrac{dy}{dx}  = (2)4x^{2-1} + (1)6x^{1-1}

\sf \rightarrow \dfrac{dy}{dx}  = 8x + 6

<u>Secondly find Slope at x = 0</u>

\sf \rightarrow \dfrac{dy}{dx}  = 8(0) + 6

\sf \rightarrow \dfrac{dy}{dx}  = 6      Yep so the slope(m) is 6.

                                       Recall, tangent line has the same slope.

<u>Now third step is to find coordinates</u>

At x = 0, y = 4x² + 6x ⇒ y = 4(0)² + 6(0) = 0   \star so coordinates: (0, 0)

Equation:

  • y - y₁ = m(x - x₁)
  • y - 0 = 6(x -0)
  • y = 6x
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