How do u figure this out?

[aisha]

If f(x)=x^(2)-2x-8 then what is the y-intercept of y=-f(x)+2?

I dont know how to figure this out. I have the answer but dont know how to get it.

[quasar987]

I'm assuming the "y-intercept" is referring ot the y-coordinates at which the two curves intersect each other?

In which case, it is always the same method to find the points of intersection between two functions:

1) you set them equal with one another

2) solve for x

This gives you the "x intersects". To find the y intercept, just plug in the x intercept in either of the two functions (since they are equal at those precise points) and this gives you the corresponding y-intercept.

[quasar987]

Now I'm thinking maybe the "y-intercept" refers to the points at which the curve intercept the y axis...

In this case, ok... f(x) = x^(2)-2x-8 and y = -f(x)+2. That is to say, y = -(x^(2)-2x-8)+2 = (-x^(2)+2x+8)+2 = -x^(2)+2x+10.

When does a function intercept the y axis? It is when x = 0. Just plug x = 0 in your function and it gives you the y coordinate at which the function intercept the y axis.

[Parth Dave]

Quasars got it right in his second post. The y-intercept is the point at which the curve crossing the y-axis. If it crosses the y-axis, x = 0. I think you can take it from there.

[aisha]

Quasars got it right in his second post. The y-intercept is the point at which the curve crossing the y-axis. If it crosses the y-axis, x = 0. I think you can take it from there.

I still dont get it , I tried doing that but didnt get the right answer the answer is (-2.3,0) and (4.3,0)

[Doc Al]

Did you solve the quadratic to find where y = 0?

[aisha]

ok when we factor the quadratic we get x=4 and x=-2
these are the x intercepts (4,0) and (-2,0) so I flipped them around to get the y intercept (0,4) and (0,-2) now the question says then what is the y intercept of y=-f(x)+2 so I multiplied the y coordinate by -1 and added 2 my answer was (0,-2) and (0,4) is this correct? If so is there an easier way? My answer in the lesson says (-2.3,0) and (4.3,0) is this wrong?

[Doc Al]

The answer given is correct; your factoring is wrong. (Multiply it out and see.) You'll need to use the quadratic formula.

[aisha]

Im stuck I dont know what to do where am I using the quadratic formula? With which equation ?

[Doc Al]

Im stuck I dont know what to do where am I using the quadratic formula? With which equation ?
In post #3, quasar987 gave you the expression for y as a function of x. Set y = 0, and solve the resulting quadratic.

[aisha]

In post #3, quasar987 gave you the expression for y as a function of x. Set y = 0, and solve the resulting quadratic.

ok so a=-1 b=2 and c=10?

I got x=-2.16 and x=4.16 the answer said (-2.3,0) and (4.3,0) where did I go wrong? And how come we are solving for x intercepts when the question asked for y-intercepts. :confused:

[apchemstudent]

In post #3, quasar987 gave you the expression for y as a function of x. Set y = 0, and solve the resulting quadratic.

why do we need to solve for the x intercepts? Dont we just need to know the y intercept?

f(x) = x^2-2x-8

we need to find y intercept of y = -f(x) + 2

plug in f(x) and as mentioned above, just plug in 0 for x to solve for y intercept. I don't see why the x intercept is needed here? Or did aisha mistake y intercepts for x intercepts?

[apchemstudent]

ok so a=-1 b=2 and c=10?

I got x=-2.16 and x=4.16 the answer said (-2.3,0) and (4.3,0) where did I go wrong? And how come we are solving for x intercepts when the question asked for y-intercepts. :confused:

you tell me...

[Doc Al]

ok so a=-1 b=2 and c=10?
Right.
I got x=-2.16 and x=4.16 the answer said (-2.3,0) and (4.3,0) where did I go wrong?
Be more careful with your arithmetic!
And how come we are solving for x intercepts when the question asked for y-intercepts.
Good point! The answers are the x-intercepts, so I just assumed that you (or the question) mixed things up. :smile:

x-intercepts are the points where y=0; y-intercepts are where x=0. (The y-intercept of this function is (0, 10).)