Why are there certain question where you multiply to a power and some you don't. Like tree diagrams. This is for probability of
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The transformed function is G(x) = -4x² after applying the transformation stretched vertically and flipped over the x-axis option (C) G(x) = -4x² is correct.
<h3>What is a function?</h3>It is defined as a special type of relationship, and they have a predefined domain and range according to the function every value in the domain is related to exactly one value in the range.
The options are missing.
The options are:
A. G(x) = 4x²
B. G(x) = -(1/4)x²
C. G(x) = -4x²
D. G(x) = (1/4)x²
We have an equation of a function F(x)
F(x) = x²
The transformation F(x) can be stretched vertically and flipped over the x-axis to produce the graph of G(x)
To stretch vertically if the function is multiplied by a constant value
f(x) = ax²
To flip over the x-axis if multiply by negative value.
g(x) = -ax²
From the options
G(x) = -4x²
Thus, the transformed function is G(x) = -4x² after applying the transformation stretched vertically and flipped over the x-axis option (C) G(x) = -4x² is correct.
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Answer:
The correct option is;
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;
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;
Which can be simplified as follows;
And further simplified as follows;
Interchanging x and y in the function of the inverse, m⁻¹(x), we have;
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;
to increase m⁻¹(x) above the minimum.