Question

y = –2x + 5

Answer 1

Yes... what is your question about y = -2x + 5?

Answer 2

yeah

Answer 3

Thats a nice line you got there

Answer 4

Since you already have it set to y-intercept form, you can easily identify the slope and y-intercept. The value attached to the x is the slope, and the value not attached to the x is the y-intercept. Therefore your slope is -2 and your y-intercept is 5.

Answer 5

This equation is linear.
The domain and range of the function is the entire set of Real numbers.
The slope of the line is -2.
The y-intercept of the line is (0,5).
The x-intercept of the line is where y = 0. (5/2, 0)
\[ \begin{split}
y &= -2x + 5\\
0 &= -2x + 5\\
-5 &= -2x\\
\frac{5}{2} &= x\\
\end{split}
\]
The inverse of y = -2x + 5 is when we switch the x and y, and solve for y. y = (-1/2)x + 5/2
\[ \begin{split}
y &= -2x + 5\\
x &= -2y + 5\\
x - 5 &= -2y\\
\frac{x-5}{-2} &= y\\
y &= \frac{x-5}{-2}\\
& = \frac{-1}{2}x + \frac{5}{2}\\
\end{split}
\]
The derivative of y is equal to the slope of y. y' = -2
The second derivative and higher are 0.
The line has no maximum or minimum.
The line has no critical points.
The indefinite integral of y is given by:
\[ \begin{split}
\int\limits_{}^{}y dx &= \int\limits_{}^{}(-2x + 5) dx\\
&= -2\int\limits_{}^{}x dx + 5\int\limits_{}^{}1 dx\\
&= -2\frac{x^{2}}{2} + 5x + C\\
&= -x^{2} + 5x + C\\
\end{split}
\]