Still another problem I missed!
- Thread starter Belby
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For any point (x1,y1) on the x-axis, which one do we know the value of, x1 or y1???
The answer is y1.
It is y that must be set to zero, not x.
PS Start sketching these problems on a graph. It is extremely enlightening! Make an x-y chart with a few points in it and plot those points. You'll be amazed how helpful this can be in understanding these problems.
D
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Draw a sketch while studying WJM's answer - it will be clear as glass.Belby said:
mmm4444bot
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Hi there, Belby:
Perhaps, we should back up to the definition of xy-coordinates!
Please carefully read through the following statements, and then I will talk about intercepts at the very end.
(x, y) are coordinates of a point.
We call this an ordered pair of numbers because the order in which the two coordinates are listed is CRUCIAL.
The x-coordinate is ALWAYS listed first.
The point (4, 7) indicates that the x-coordinate is 4 and the y-coordinate is 7.
DO NOT WRITE (7, 4) because this address describes an entirely different point.
Next, what do these coordinates mean? (I'm going to assume that you know how to draw the x-axis and y-axis; in other words, I'm assuming that you know the x-axis is the HORIZONTAL axis and the y-axis is the VERTICAL axis.)
The x-coordinate is a number, and this number represents a DISTANCE. Specifically, the x-coordinate gives us the HORIZONTAL distance away from the Y-AXIS.
(4, 7) shows that the x-coordinate is 4. This means that the point is located a horizontal distance of 4 units away from the y-axis (to the right).
(-2, -9) shows that the x-coordinate is -2. This means that the point is located a horiztonal distance of 2 units away from the y-axis (to the left).
NOTE! In these two examples, neither 4 nor -2 says anything about the VERTICAL placement of the point. We must use the corresponding y-coordinate to show how far up or down the point lies.
So, the y-coordinate is also a number, and this number also represents a DISTANCE. Specifically, the y-coordinate gives us the VERTICAL distance away from the X-AXIS.
(4, 7) shows that the y-coordinate is 7. This means that the point is located a vertical distance of 7 units away from the x-axis (up).
(-2, -9) shows that the y-coordinate is -9. This means that the point is located a vertical distance of 9 units away from the x-axis (down).
NOTE! In these two examples, neither 7 nor -9 says anything about the HORIZONTAL placement of the point. We must use the corresponding x-coordinate to show how far right or left the point lies.
Okay, if this is clear, so far, then the following should be clear as glass.
The terminology "x-intercept" describes some point that lies ON THE X-AXIS.
The terminology "y-intercept" describes some point that lies ON THE Y-AXIS.
Let's examine in greater detail what we can deduce about any x-intercept.
Since x-intercepts are located ON THE X-AXIS itself, they never lie above or below the x-axis. That's obvious, right? Well, if they lie directly on the axis, then their vertical distance away from the x-axis (up or down) MUST BE ZERO. Always!
In other words, the y-coordinate of all x-intercepts is ZERO because (as I described above) the y-coordinate IS the vertical distance away from the axis.
(4, 0) (-17, 0) (22/7, 0) (-23.478, 0)
All of these points lie ON the x-axis because the y-coordinate of each is 0. That zero tells us not to move ANY distance above or below the x-axis, when we plot the point.
Likewise, all points that lie on the y-axis have x-coordinates of zero because points on the y-axis are not located ANY horizontal distance away (either to the left or right).
(0, 7) (0, -23.5) (0, 4/5) (0, 4444)
All of these point lie ON the y-axis because the x-coordinate of each is 0. That zero tells us not to move ANY distance to the left or right of the y-axis, when we plot the point.
So, what does all of this have to do with the exercise that you posted in this thread? I'll tell you.
Here's your exercise: "Find the points where the graph of y + 3 = (x - 3)^2 crosses the x-axis."
The phrase "points where graph crosses the x-axis" specifically refers to X-INTERCEPTS.
Therefore, you know right away -- without doing any calculations at all -- that the y-coordinate of these intersection points with the x-axis MUST BE ZERO.
In other words, you're looking for points of the form (x, 0) because (and I'm hoping that you realize this now) the y-coordinate of ANY x-intercept must be 0. The point where a graph intersects the x-axis is actually ON THE X-AXIS, so it's distance away from the x-axis is zero.
This is why you set y = 0 when searching for x-intercepts.
Guess what? To find y-intercepts (i.e., the point where a graph crosses the vertical axis), then we know the coordinates take the form (0,y). So, we would set x = 0, and go from there.
If your glass is still dirty, then let me know. I can upload some pictures, if you think that would help.
And, as always, if I wrote anything that you don't understand, then ask specific questions. (I do make several mistakes each decade. :wink: )
Cheers ~ Mark
Mark,
Every answer I have input into the system is wrong according to the computer. Now the problem reads find the points where the graph of y+3=(x-3)2 crosses the x-axis. The x-coordinate of the right most point is x=?
Is the 2 the same as ^2 because this is what the computer now says for the problem and every answer I have put in is wrong. Should I put in points (3,0) or just 3. Help one more time, please.
Every answer I have input into the system is wrong according to the computer. Now the problem reads find the points where the graph of y+3=(x-3)2 crosses the x-axis. The x-coordinate of the right most point is x=?
Is the 2 the same as ^2 because this is what the computer now says for the problem and every answer I have put in is wrong. Should I put in points (3,0) or just 3. Help one more time, please.
mmm4444bot
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Belby said:
Yes, the 2 is supposed to be an exponent because the exercise states that there are "points" (i.e., plural), and you're supposed to look for the "rightmost" point. So, we know that the 2 is not a factor. Otherwise, the graph would be a straight line instead of a parabola. In other words, there would be only ONE x-intercept. Turns out to be x = 9/2.
The 2 is an exponent. The given equation is quadratic.
How did you get x = 3 when y = 0 ?
Did you use the Quadratic Formula ?
Belby said:
Mark has given you a WONDERFUL explanation about what coordinates mean, how to find the location of a point corresponding to a particular ordered pair, and how to determine SOMETHING about the coordinates of a point that is on either the x-axis or the y-axis.
I assume that the "2" you've written on
y + 3 = (x - 3)2
is really an exponent. And we indicate exponents using " ^ " before the exponent, or using one of the advanced formatting features, such as the "sup" button.
y + 3 = (x - 3)^2
Now...if you are looking for a point on the x-axis, and if you have carefully read Mark's great explanation, you should realize that if a point is on the x-axis, the y-coordinate of that point should be 0.
So....you know y = 0 at the point(s) you're looking for. Substitute 0 for y:
y + 3 = (x - 3)^2
0 + 3 = (x - 3)^2
3 = (x - 3)^2
Can you solve this for x?
Take the square root of both sides to eliminate the "squared"
+ sqrt(3) = sqrt[(x - 3)^2]
+ sqrt(3) = x - 3
Now...add 3 to both sides of the equation....
And remember, you're looking for the RIGHT-MOST point of intersection on the x-axis...that will be the largest of the two values of x.