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sleet_krkn [62]
3 years ago
8

Write the equation of a line perpendicular to y=-2x+1 through (2,3) in slope intercept form.

Mathematics
1 answer:
Fed [463]3 years ago
7 0

Answer:

y=\dfrac{1}{2}x+2

Step-by-step explanation:

y= - 2x + 1

Compare with y = mx + b

m = Slope = -2

Slope of the perpendicular line = -1/m = -1/-2 = 1/2

Point (2,3)

Equation : y -y1 =m(x -x1)

y - 3 = \dfrac{1}{2}*(x - 2)\\\\\\y-3=\dfrac{1}{2}*x - 2*\dfrac{1}{2}\\\\\\y -3=\dfrac{1}{2}x-1\\\\y =\dfrac{1}{2}x-1+3\\\\y=\dfrac{1}{2}x+2

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PLEASE ANSWER ASAP!!
Rufina [12.5K]

Answer:

  1. The instantaneous rate of increase of f(x) at x_{0} = 3 is 3.
  2. One possible equation of this line is y = 3x - 16.
  3. The line is tangent to the graph of y = f(x). The slope of the line is the same as the instantaneous rate of increase of f(x) at x_{0} = 3.

Step-by-step explanation:

<h3>1.</h3>

\begin{aligned}&\lim_{h\to 0}{\frac{f(3+h)-f(h)}{h}}\\ =&\lim_{h\to 0}\frac{1}{h} \cdot [(3^{3} + 3 \times 3^{2}h +3\times 3h^{2} + h^{3})- 4(3^{2} + 2\times 3h + h^{2}) + 2 \\[-1em] &\phantom{\lim_{h\to 0}\frac{1}{h}\cdot []} -(3^{3} -4\times 3^{2} + 2)]\\ = &\lim_{h \to 0}\frac{h^{3} + 5h^{2} +3h}{h}\\=& \lim_{h \to 0}{h^{2}} + \lim_{h \to 0}{5h} + \lim_{h\to 0}{3}\\ =& 3\end{aligned}.

<h3>2.</h3>

f(3) = 3^{3} - 4\times 3^{2} + 2 =-7.

In other words, the graph of y = f(x) passes through the point (3, -7) where x = 3.

The point-slope form of a line in a cartesian plane is:

y - y_0 = m (x - x_0).

For this line,

(3, -7) is the point on the line, while

3 is the slope of the line.

The equation of this line will thus be

y - (-7) = 3(x - 3).

That's equivalent to

y = 3x - 16.

<h3>3.</h3>

Refer to the diagram attached. The line touches the graph of y = f(x) at x = 3 without crossing it. The line here is thus a tangent to the graph of y = f(x) at x = 3. The slope of the line represents the instantaneous rate of increase of f(x) at x_{0} = 3.

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4×+2.5=-1

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x=0.875

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