Parabola problem
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Otis
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Hello! My reply uses the exercise labels as given.
In part (i), they say the y-intercept is the height of the ledge. Actually, it's the y-coordinate of the y-intercept which gives the height of the ledge, but I understand that many people blur the distinction and simply say the y-coordinate itself is the point of intersection. (Personally, I try to not do that anymore.)
If I were answering part (i), I would report the y-intercept as a point: (0, 1.2)
If they had asked for the height of the ledge, then I would include units and report 1.2 meters
In part (ii)(1), they ask for coordinates of the point on the graph where x=5.
Your first attempt is the right approach (evaluating y when x is 5), but you began with
0 = -0.15(5)^2 + (5) + 1.2
Only the right-hand side is correct (substituting 5 for x). You're trying to find y, so don't set it equal to zero:
y = -0.15(5)^2 + (5) + 1.2
Your evaluation is correct, but the answer isn't y=2.45 because they've asked for coordinates of a point. Report (5, 2.45).
Your second attempt -- using the quadratic formula -- is the wrong approach. The quadratic formula yields x-values, but we already know x=5.
For part (ii)(2), your answer should be 'yes' instead of 'no'.
When the kangaroo is five meters horizontally from the ledge, its horizontal location is at x=5. The result in part (ii)(1) tells us the kangaroo's feet are 2.45 meters above the ground, at that location. The obstacle is only 1 meter tall, so the kangaroo clears it.
Part (iii)(1) seems to ask for the x-intercepts of the entire parabola. That is, they're no longer talking about the model because there are no negative values of x in the model. (They've implied that x-values start at the jump, and we have no idea how far the ledge extends to the left.)
So, you're probably right to report both x-intercepts. However, you forgot to type the negative sign when reporting -1.03829235 to two significant figures. (Also, some of the radical signs are not correctly formatted.) Your initial substitutions in the quadratic formula are good and your results are correct, so I didn't verify the intermediate steps.
Because x-intercepts are points, I would report the coordinates and omit units:
(-1.0, 0) and (7.7, 0)
Part (iii)(2) switches back to the model, and it asks for a distance (so we include the units). You didn't type your answer for part (iii)(2), but I'm sure you intended to write 7.7 meters. Correct!
PS: If your class actually applies rules of significant figures (versus simple rounding) to such word problems, then I would've expected to see the model given as:
y = -0.15x^2 + 1.0x + 1.2
Cheers ?
In part (i), they say the y-intercept is the height of the ledge. Actually, it's the y-coordinate of the y-intercept which gives the height of the ledge, but I understand that many people blur the distinction and simply say the y-coordinate itself is the point of intersection. (Personally, I try to not do that anymore.)
If I were answering part (i), I would report the y-intercept as a point: (0, 1.2)
If they had asked for the height of the ledge, then I would include units and report 1.2 meters
In part (ii)(1), they ask for coordinates of the point on the graph where x=5.
Your first attempt is the right approach (evaluating y when x is 5), but you began with
0 = -0.15(5)^2 + (5) + 1.2
Only the right-hand side is correct (substituting 5 for x). You're trying to find y, so don't set it equal to zero:
y = -0.15(5)^2 + (5) + 1.2
Your evaluation is correct, but the answer isn't y=2.45 because they've asked for coordinates of a point. Report (5, 2.45).
Your second attempt -- using the quadratic formula -- is the wrong approach. The quadratic formula yields x-values, but we already know x=5.
For part (ii)(2), your answer should be 'yes' instead of 'no'.
When the kangaroo is five meters horizontally from the ledge, its horizontal location is at x=5. The result in part (ii)(1) tells us the kangaroo's feet are 2.45 meters above the ground, at that location. The obstacle is only 1 meter tall, so the kangaroo clears it.
Part (iii)(1) seems to ask for the x-intercepts of the entire parabola. That is, they're no longer talking about the model because there are no negative values of x in the model. (They've implied that x-values start at the jump, and we have no idea how far the ledge extends to the left.)
So, you're probably right to report both x-intercepts. However, you forgot to type the negative sign when reporting -1.03829235 to two significant figures. (Also, some of the radical signs are not correctly formatted.) Your initial substitutions in the quadratic formula are good and your results are correct, so I didn't verify the intermediate steps.
Because x-intercepts are points, I would report the coordinates and omit units:
(-1.0, 0) and (7.7, 0)
Part (iii)(2) switches back to the model, and it asks for a distance (so we include the units). You didn't type your answer for part (iii)(2), but I'm sure you intended to write 7.7 meters. Correct!
PS: If your class actually applies rules of significant figures (versus simple rounding) to such word problems, then I would've expected to see the model given as:
y = -0.15x^2 + 1.0x + 1.2
Cheers ?