Here i how I would do it:f(x)=−x2+8x+15
set f(x) = 0 to find the points at which the graph crosses the x-axis. So−x2+8x+15=0
multiply through by -1x2−8x−15=0 (x−4)2−31=0 x=4±31−−√
So these are the points at which the graph crosses the x-axis. To find the point where it crosses the y-axis, set x=0 in your original equation to get 15. Now because of the negative on the x^2, your graph will be an upside down parabola, going through(0,15),(4−31−−√,0)and(4+31−−√,0)
To find the coordinates of the maximum (it is maximum) of the graph, you take a look at the completed square method above. Since we multiplied through by -1, we need to multiply through by it again to get:f(x)=31−(x−4)2
Now this is maximal when x=4, because x=4 causes -(x-4)^2 to vanish. So the coordinates of the maximum are (4,y). To find the y, simply substitute x=4 into the equation f(x) to give y = 31. So it agrees with the mighty Satellite: (4,31) is the vertex.

8 0

3 years ago

NEED HELP ASAP PLEASE!!

qwelly [4]

Option D:

$$y=\frac{5x-2}{x}$

Solution:

Given function:

$$y=\frac{-2}{x-5}$

To find the inverse of the function.

Inverse of a function:

If a function f(x) is mapping x to y, then the inverse function of f(x) maps y back to x.

$$y=\frac{-2}{x-5}$

Interchange the variables x and y.

$$x=\frac{-2}{y-5}$

Now, solve for y.

Multiply both sides by (y - 5).

$$x(y-5)=-\frac{2}{y-5}(y-5)$

Cancel the common factors, we get

$x(y-5)=-2$

Divide by x on both sides.

$$y-5=-\frac{2}{x}$

Add 5 on both sides.

$$y=-\frac{2}{x}+5$

$$y=-\frac{2}{x}+\frac{5}{1}$

To make the denominator same, multiply the 2nd term by $\frac{x}{x}$ .

$$y=-\frac{2}{x}+\frac{5x}{x}$

$$y=\frac{-2+5x}{x}$

$$y=\frac{5x-2}{x}$

Option D is the correct answer.

8 0

2 years ago

If men having High Blood Sugar problems are diagnosed with Diabetes, with the mean blood sugar level to be at 150 and a standard

frutty [35]

Answer:

0.0062

Step-by-step explanation:

Let X be the diagnosed blood sugar level.

Probability of committing a Type II error = Probability of failing to detect the disease when the person has the disease = Probability that the diagnosed blood sugar level is less than or equal to 125 =

$P(X \le 125 |\mu = 150)$

$= P[Z\le (125 - 150)/10]$

$=P[Z \le -2.5]$

= 0.0062

8 0

2 years ago