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3 years ago

Below are the functions of y=|x| and y=|x|-5 how are the functions related.

Mathematics

1 answer:

natita [175]3 years ago

7 0

Answer:

Option D is correct.

The functions have the same shape.

The y-intercepts of y=|x| is 0, and the y-intercepts of the second function y=|x|-5 is, -5

Step-by-step explanation:

Modulus function state that a function which gives a positive output irrespective of the input.

The graph of the functions as shown below in the attachment :

The functions are; $y= |x| -5$ makes a "V'" shape much like y=|x|.

Every absolute value graph will take this same shape, as long as the expression inside the absolute value is linear.

therefore, these functions $y= |x|$ and $y= |x| -5$ have the same shape.

y-intercepts defined as the graph crosses the y-axis

i.e substitute the value of x=0 and solve for y.

For the function y = |x|

to find the y-intercepts:

substitute value of x =0 we get;

y = |0| = 0

Then, the y-intercept of the function y=|x| is 0.

Similarly, for the function y = |x|-5

Find: y-intercept

substitute the value of x =0 we get;

y = |0| - 5 = -5

Therefore, the y-intercept of the function y=|x|-5 is -5

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stealth61 [152]

Answer:

The correct option is;

$The \ domain \ restriction \ x \geq \dfrac{17}{2} \ results \ in \ m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }}$

Step-by-step explanation:

The given information is that m(x) = x² - 17·x

The above equation can be written in the form;

y = x² - 17·x

Therefore;

0 = x² - 17·x - y

From the general solution of a quadratic equation, 0 = a·x² + b·x + c we have;

$x = \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}$

By comparison to the equation,0 = x² - 17·x - y, we have;

a = 1, b = -17, and c = -y

Substituting the values of a, b and c into the formula for the general solution of a quadratic equation, we have;

$x = \dfrac{-(-17)\pm \sqrt{(-17)^{2}-4\times (1) \times (-y)}}{2\times (1)} = \dfrac{17\pm \sqrt{289+4\cdot y}}{2}$

Which can be simplified as follows;

$x = \dfrac{17\pm \sqrt{289+4\cdot y}}{2}= \dfrac{17}{2} \pm \dfrac{1}{2} \times \sqrt{289+4\cdot y}} = \dfrac{17}{2} \pm \sqrt{\dfrac{289}{4} +\dfrac{4\cdot y}{4} }}$

And further simplified as follows;

$x = \dfrac{17}{2} \pm \sqrt{\dfrac{289}{4} +y }} = \dfrac{17}{2} \pm \sqrt{y + \dfrac{289}{4} }}$

Interchanging x and y in the function of the inverse, m⁻¹(x), we have;

$m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }}$

We note that the maximum or minimum point of the function, m(x) = x² - 17·x found by differentiating the function and equating the result to zero, gives;

m'(x) = 2·x - 17 = 0

x = 17/2

Similarly, the second derivative is taken to determine if the given point is a maximum or minimum point as follows;

m''(x) = 2 > 0, therefore, the point is a minimum point on the graph

Therefore, as x increases past the minimum point of 17/2, m⁻¹(x) increases to give;

$The \ domain \ restriction \ x \geq \dfrac{17}{2} \ results \ in \ m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }}$ to increase m⁻¹(x) above the minimum.

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