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denis-greek [22]
3 years ago
13

What is the equation of a line, in general form, that passes though points (-1,2) and (5,2)?

Mathematics
1 answer:
Tom [10]3 years ago
3 0

Answer:

A

Step-by-step explanation:

To write the equation of the line. first calculate the slope using the slope formula.

m = \frac{y_2-y_1}{x_2-x_1} = \frac{2-2}{5--1}= \frac{0}{6} = 0

Since the slope of this line is 0, it is horizontal and has the form y=b where b is the y-coordinate. So y = 2 is the equation. In general form, it would be y-2 = 0

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A. 0

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Consider the function: f(x) = 2x^2 - 1.5x - 5 and its graph.
Dominik [7]

Answer:

f(x) = -4x² + 19x - 18

Step-by-step explanation:

f(x) = 2x^2 - 1.5x - 5

i) If it is translated 2 units in the positive x direction, therefore we use f(x-2)

y=f(x-2) = 2(x-2)^2 - 1.5(x-2) - 5\\y = 2(x^2-4x+4) - 1.5x+3 - 5\\y=2x^2-8x+8-1.5x+3-5\\y=2x^2-9.5x+6

f(x) = 2x² - 9.5x + 6

If it is then translated 3 units in the positive y, we add 3 to the input function to get:

y=f(x)+3=2x^2-9.5x+6+3\\y = 2x^2-9.5x+9

ii) stretched vertically by a factor of 2, we multiply the function by 2 to get:

y =2( 2x^2-9.5x+9) = 4x^2-19x+18\\y= 4x^2-19x+18

iii) reflected across the x-axis

we multiply the parent function by –1, to get a reflection about the x-axis.

y= -1(4x^2-19x+18)=-4x^2+19x-18\\y=f(x)=-4x^2+19x-18

5 0
3 years ago
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