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3 years ago

Can somebody please help meee out ?? I don’t understand this .

Mathematics

2 answers:

borishaifa [10]3 years ago

6 0

Answer:

(-5,3)-3/2=(-1,-3)

Step-by-step explanation:

So the slope intercept form is 3/2 and the points are (-5,3) and (-1,-3)

y intercept form is y=mx+b.

I hope one of those helps

slavikrds [6]3 years ago

5 0

Answer:

the equation for the function shown is y= -3/2x -9/2

Step-by-step explanation:

You can find the slope of the graph by counting how many spaces up or down over how many spaces left or right between two points. Start from the left of the graph and go towards the right of the graph when doing this. If you count the spaces between the two points on the graph, it is down six units (-6 on top) and 4 units right (4 on bottom) to get a slope of -6/4, which can be simplified by dividing both the top and bottom by two to get a slope of -3/2. Next you can use the point-slope form to find the equation of the graph in the form y=mx + b. Point slope form looks like this: y-y1=(the slope)(x-x1).

Plug in what you got for the slope and the coordinates for any point on the graph (the y coordinate replacing y1 and the x coordinate replacing x1). Here, I used the point (-1, -3), but you should get the same answer no matter what point you use, as long as it is a point on the graph:=> y-(-3)= -3/2(x-(-1))

=>y+3=-3/2(x+1)

distribute the -3/2 to both terms in the parentheses: y+3= -3/2x -3/2

subtract 3 from both sides to get y by itself (aka to convert to the form y=mx+b): y= -3/2x -9/2 (Here, -9/2 is your y-intercept, or where the graph crosses the y-axis).

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2 years ago

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I have calculus problems that I need help with.

aleksklad [387]

a. Note that $f(x)=x^ne^{-2x}$ is continuous for all $x$ . If $f(x)$ attains a maximum at $x=3$ , then $f'(3) = 0$ . Compute the derivative of $f$ .

$f'(x) = nx^{n-1} e^{-2x} - 2x^n e^{-2x}$

Evaluate this at $x=3$ and solve for $n$ .

$n\cdot3^{n-1} e^{-6} - 2\cdot3^n e^{-6} = 0$

$n\cdot3^{n-1} = 2\cdot3^n$

$\dfrac n2 = \dfrac{3^n}{3^{n-1}}$

$\dfrac n2 = 3 \implies \boxed{n=6}$

To ensure that a maximum is reached for this value of $n$ , we need to check the sign of the second derivative at this critical point.

$f(x) = x^6 e^{-2x} \\\\ \implies f'(x) = 6x^5 e^{-2x} - 2x^6 e^{-2x} \\\\ \implies f''(x) = 30x^4 e^{-2x} - 24x^5 e^{-2x} + 4x^6 e^{-2x} \\\\ \implies f''(3) = -\dfrac{486}{e^6} < 0$

The second derivative at $x=3$ is negative, which indicate the function is concave downward, which in turn means that $f(3)$ is indeed a (local) maximum.

b. When $n=4$ , we have derivatives

$f(x) = x^4 e^{-2x} \\\\ \implies f'(x) = 4x^3 e^{-2x} - 2x^4 e^{-2x} \\\\ \implies f''(x) = 12x^2 e^{-2x} - 16x^3e^{-2x} + 4x^4e^{-2x}$

Inflection points can occur where the second derivative vanishes.

$12x^2 e^{-2x} - 16x^3 e^{-2x} + 4x^4 e^{-2x} = 0$

$12x^2 - 16x^3 + 4x^4 = 0$

$4x^2 (3 - 4x + x^2) = 0$

$4x^2 (x - 3) (x - 1) = 0$

Then we have three possible inflection points when $x=0$ , $x=1$ , or $x=3$ .

To decide which are actually inflection points, check the sign of $f''$ in each of the intervals $(-\infty,0)$ , $(0, 1)$ , $(1, 3)$ , and $(3,\infty)$ . It's enough to check the sign of any test value of $x$ from each interval.