The point-slope form of the equation for a line can be written as
... y = m(x -h) +k . . . . . . . for a line with slope m through point (h, k)
Your function gives
... f'(h) = m
... f(h) = k
a) The tangent line is then
... y = 5(x -2) +3
b) The normal line will have a slope that is the negative reciprocal of that of the tangent line.
... y = (-1/5)(x -2) +3
_____
You asked for "an equation." That's what is provided above. Each can be rearranged to whatever form you like.
In standard form, the tangent line's equation is 5x -y = 7. The normal line's equation is x +5y = 17.
Answer:
I guess 21 is your answers.
Answer:
g=3
Step-by-step explanation:
9 - g = 2g
+g +g
9=3g
divide both sides by 3
g=9
It's D: (0,0)
This is because when you put 0 in for x your y value will equal 0 and that is one of your answers, so it's the point (0,0)
