on number 3 y=kx
so would i divide y by x and if so I get 1.57, 1.62, 1.66, and 1.7..Is this right?
I have a few questions that I don't understand hope someone can help.
1)write an equation of th line, in point slope form, that passes through the two given points. (-17,8),(3,-2)
I figured (y-8)=-2(x+17) not sure
2) the graph of g(x) is f(x) vertically compressed by a factor of 1/7 and reflected in the x-axis what is the function rule for g(x)given f(x)=2x.
not sure about this one
3)determine whether y varies directly with x if so find the constant of variation k and write the equation.
x y
7 11
8 13
9 15
10 17
I know you start with y=kx but then I'm lost.
not looking for just answers but if someone can explain I'd appreciate it
2 answers
#1. If not sure, plug in the points and see whether they satisfy your equation. you will see that they don't even come close
The (y-8) and (x+17) bits are ok, but how did you come up with a slope of -2? (-2-8)/(3+17) = -1/2
Looks like you used ∆x/∆y for the slope, rather than ∆y/∆x!
#2: you want g(x) which produces a height 1/7 of f(x). So,
g(x) = 1/7 f(x)
Now, you want to reflect g(x) around the x-axis, so
h(x) = -g(x)
h(x) = -1/7 f(x) = -1/7 (2x) = -2/7 x
check: f(1) = 2
h(1) = -2/7 shrunk, reflected
that is correct. Since the ratio changes, it is not a constant k.
y does not vary directly with x
In fact, y increases by 2 for every increase in x by 1. So,
y = 2x-3
That pesky -3 fouls up the direct variation.
The (y-8) and (x+17) bits are ok, but how did you come up with a slope of -2? (-2-8)/(3+17) = -1/2
Looks like you used ∆x/∆y for the slope, rather than ∆y/∆x!
#2: you want g(x) which produces a height 1/7 of f(x). So,
g(x) = 1/7 f(x)
Now, you want to reflect g(x) around the x-axis, so
h(x) = -g(x)
h(x) = -1/7 f(x) = -1/7 (2x) = -2/7 x
check: f(1) = 2
h(1) = -2/7 shrunk, reflected
that is correct. Since the ratio changes, it is not a constant k.
y does not vary directly with x
In fact, y increases by 2 for every increase in x by 1. So,
y = 2x-3
That pesky -3 fouls up the direct variation.
2 answers