Answer:
do u need to solve for x or for y ?
Step-by-step explanation:
F(x)=x+c, where c is an arbitrary constant.
if c is positive then translation above
if c is negative then translation down
reflection of f(x)=x^2 across x-axis then
f(x)=-x^2
keeping in mind that perpendicular lines have negative reciprocal slopes, hmmm what's the slope of the equation above anyway?
![\bf y = \cfrac{2}{3}x\implies y = \stackrel{\stackrel{m}{\downarrow }}{\cfrac{2}{3}}x+0\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20y%20%3D%20%5Ccfrac%7B2%7D%7B3%7Dx%5Cimplies%20y%20%3D%20%5Cstackrel%7B%5Cstackrel%7Bm%7D%7B%5Cdownarrow%20%7D%7D%7B%5Ccfrac%7B2%7D%7B3%7D%7Dx%2B0%5Cqquad%20%5Cimpliedby%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20slope-intercept~form%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y%3D%5Cunderset%7By-intercept%7D%7B%5Cstackrel%7Bslope%5Cqquad%20%7D%7B%5Cstackrel%7B%5Cdownarrow%20%7D%7Bm%7Dx%2B%5Cunderset%7B%5Cuparrow%20%7D%7Bb%7D%7D%7D%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
so we're really looking for the equation of a line whose slope is -3/2 and runs through (0,0).
Simplify \frac{5}{3}x35x to \frac{5x}{3}35x
x-\frac{5x}{3}<3x−35x<3
2
Simplify x-\frac{5x}{3}x−35x to -\frac{2x}{3}−32x
-\frac{2x}{3}<3−32x<3
3
Multiply both sides by 33
-2x<3\times 3−2x<3×3
4
Simplify 3\times 33×3 to 99
-2x<9−2x<9
5
Divide both sides by -2−2
x>-\frac{9}{2}x>−29
Answer:
Option ( A ) is the answer at (2, 2), (3, 1), (4, 2)
Step-by-step explanation:
