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3 years ago

Find the vertex of this function. Is it a maximum or a minimum? y=-3(x + 4)2 + 2

Mathematics

1 answer:

Vika [28.1K]3 years ago

8 0

Step-by-step explanation:

The equation given is of the form

$y = a(x - h)^2 + k$

where (h, k) are the coordinates of the vertex. So in this case, the vertex is located at (4, 2). Since a is a negative number (a = -3), this means that the vertex is a maximum because the parabola opens downward.

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3 years ago

The graph of an inverse trigonometric function passes through the point (1, pi/2). Which of the following could be the equation

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Step-by-step explanation:

Since, the graph of an inverse trigonometric function will pass through the point $(1,\frac{\pi}{2})$ ,

If this point satisfies the function,

For the function $y=cos^{-1} x$

If x = 1

$y=cos^{-1}1=0$

Thus, $(1,\frac{\pi}{2})$ is not satisfying function $y=cos^{-1} x$ ,

⇒ The graph of $y=cos^{-1} x$ is not passing through the point $(1,\frac{\pi}{2})$

For the function $y=cot^{-1}x$

If x = 1

$y=cot^{-1}1=\frac{\pi}{4}$

Thus, $(1,\frac{\pi}{2})$ is not satisfying function $y=cot^{-1}x$ ,

⇒ The graph of $y=cot^{-1}x$ is not passing through the point $(1,\frac{\pi}{2})$

For the function $y=sin^{-1} x$

If x = 1

$y=sin^{-1}1=\frac{\pi}{2}$

Thus, $(1,\frac{\pi}{2})$ is satisfying function $y=sin^{-1} x$ ,

⇒ The graph of $y=sin^{-1} x$ is passing through the point $(1,\frac{\pi}{2})$ .

For the function $y=tan^{-1}x$

If x = 1

$y=tan^{-1}1=\frac{\pi}{4}$

Thus, $(1,\frac{\pi}{2})$ is not satisfying function $y=cos^{-1} x$ ,

⇒ The graph of $y=tan^{-1} x$ is not passing through the point $(1,\frac{\pi}{2})$ .

Hence, Option C is correct.

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3 years ago