Answer:
To a power, elevated to something
One A
y = e^x
dy/dx = e^x The f(x) = the differentiated function. Any value that e^x can have, the derivative has the same value. x is contained in all the reals.
One B
y = x*e^x
y' = e^x + xe^x Using the multiplication rule.
You want the slope and the value of the of y to be the same. The slope is y' of the tangent line
xe^x = e^x + xe^x
e^x = 0
This happens only when x is very "small" like x = - 4444444
y = e^x * ln(x) Using the multiplication rule again, we need the slope of the line with is y'
y1 = e^x
y1' = e^x
y2 = ln(x)
y2' = 1/x
y' = e^x*ln(x) + e^x/x So at x = 1 the slope of the line =
y' = e^1*ln(1) + e^1/1
y' = e*0+e = e
y = mx + b
y = ex + b
to find b we use y= e^x ln(x)
e^x ln(x) = e*x + b
e^1 ln(1) = e*1 + b
ln(1) = 0
0 = e + b
b = - e
line equation and answer.
y = e*x - e
Answer:
the answer to your question is A
Answer:
What is the point used in the equation of the line y+4=1/2(x-2)
The other format for straight-line equations is called the "point-slope" form. For this one, they give you a point (x1, y1) and a slope m, and have you plug it into this formula:
y - y1 = m(x - x1)
Don't let the subscripts scare you. They are just intended to indicate the point they give you. You have the generic "x" and generic "y" that are always in your equation, and then you have the specific x and y from the point they gave you; the specific x and y are what is subscripted in the formula. Here's how you use the point-slope formula
They've given me m = 4, x1 = -1, and y1 = -6. I'll plug these values into the point-slope form, and solve for "y=":
y - y1 = m(x - x1)
y - (-6) = (4)(x - (-1))
y + 6 = 4(x + 1)
y + 6 = 4x + 4
y = 4x + 4 - 6
y = 4x - 2
Yes it can, it has to be a fraction
