What are the intercepts for the given quadratic function?

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Homework Help Overview

The discussion revolves around finding the intercepts of a quadratic function, specifically focusing on the function f(x) = x² - 2x - 8 and its transformation y = -f(x) + 2. Participants are trying to clarify the distinction between y-intercepts and x-intercepts.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the definition of y-intercepts and question whether it refers to points of intersection between two curves or points where the curve intersects the y-axis. There are discussions about setting equations equal to find intersections and plugging in values to find intercepts.

Discussion Status

Some participants have provided guidance on how to approach finding intercepts, while others express confusion about the distinction between y-intercepts and x-intercepts. There is acknowledgment of differing interpretations of the question, and some participants are attempting to clarify their understanding of the problem.

Contextual Notes

There is uncertainty regarding the correct interpretation of "y-intercept" in the context of the problem, leading to discussions about the need to solve for x-intercepts instead. Participants also mention specific values for intercepts, indicating a potential discrepancy in calculations.

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

I don't know how to figure this out. I have the answer but don't know how to get it.
 
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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.
 
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.
 
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.
 
Parth Dave said:
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 don't get it , I tried doing that but didnt get the right answer the answer is (-2.3,0) and (4.3,0)
 
solve the quadratic

Did you solve the quadratic to find where y = 0?
 
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?
 
The answer given is correct; your factoring is wrong. (Multiply it out and see.) You'll need to use the quadratic formula.
 
Im stuck I don't know what to do where am I using the quadratic formula? With which equation ?
 
  • #10
aisha said:
Im stuck I don't 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.
 
  • #11
Doc Al said:
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:
 
Last edited:
  • #12
Doc Al said:
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?
 
  • #13
aisha said:
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...
 
  • #14
aisha said:
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).)
 

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