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2 years ago

Let f(x)=(18)3^x. find the equation of the line between the points (-2,f(-2)) and (1,f(1))

1 answer:

6 0

$\text{Given that,} \\\\f(x) = 18 \cdot 3^x \\\\f(-2) = 18 \cdot 3^{-2} = 2\\\\f(1) = 18 \cdot 3^1 = 54\\ \\\text{So,}~ (x_1,y_1) = (x_1, f(-2)) = (-2,2) ~\text{and}~ (x_2,y_2) = (x_2, f(1)) = (1,54)\\ \\\text{Slope,}~ m = \dfrac{y_2 -y_1}{x_2 -x_1} = \dfrac{54-2}{1+2} = \dfrac{52}3 \\\\$

$\text{Equation of line,}\\\\~~~~~~~y - y_1 = m (x-x_1)\\\\\\\implies y-2 = \dfrac{52}3(x+2)~~~~~~~~~~~;[\text{Point-Slope form.]} \\\\\\\implies y-2 = \dfrac{52}3x + \dfrac{104}3\\\\\\\implies y = \dfrac{52}{3}x + \dfrac{104} 3 +2\\\\\\\implies y = \dfrac{52}3 x +\dfrac{110}3 ~~~~~~~~~~~~~~~;[\text{Slope -Intercept form}]$

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sukhopar [10]

Answer:

The required equation is:

$y = -\frac{4}{3}t^2 -\frac{8}{3}t + 4$

Explanation:

Let us assume that the hole is at y = 0m, with x as the time.

From the question we have (-1s, 8m) as the vertex (here x being the time variable is supposed to be in seconds and y being the distance variable is supposed to be in meters)

At x = 1s, the ball gets to the hole, therefore we have point (1s, 0m)

We know that the vertex of the parabola y = ax² + bx + c is at

$x =\frac{-b}{2a}$

therefore we have:

$-1 = \frac{-b}{2a}$

We then have the following equations:

$8 = a\times -1^2 + b\times -1 + c$

$0 = a\times -1^2 + b\times 1 + c$

$-1 = \frac{-b}{2a}$

From the 3rd equation we have

1 X 2a = b.

Therefore we have:

$8 = a\times -1^2 - 1\times 2a\times1 + c$

$0 = a\times 1^2 + 1 \times2a\times 1 + c$

We can simplify both equations and get:

$8 = a\times( -1^2 - 2s^2) + c = -a\times 3^2 + c$

$0 = a\times(1^2 + 2^2) + c = a\times 3^2 + c$

The first equation now becomes:

$8 = -a\times 3 - a\times 3 = -a\times 6$

$a = frac{8}{-6} = -\frac{4}{3}$

With a, we can find the values of c and b.

$c = -a\times3 = -(-\frac{4}{3})*3 = 4$

$b = 1\times 2a = 1\times 2(-\frac{4}{3})= -\frac{8}{3}$

Then the equation is:

$y = -\frac{4}{3}t^2 -\frac{8}{3}t + 4$

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