How to solve this problem of finding the Arch support?
$begingroup$
I am self-studying precalculus on my own. This is a problem from Ron Larson's precalculus 10e
The arch support of a bridge is
modeled by
$y = −0.0012x^2+
300$, where x and y are
measured in feet and the x-axis represents the ground.
(a) Use a graphing utility to graph the equation.
(b) Find one x-intercept of the graph. Explain how to
use the intercept and the symmetry of the graph to
find the width of the arch support.
For the first question (a) I graphed it on desmos
And for the second problem my approach was to use the distance formula and the answer was 1000feet. Since I am doing everything all on my own I have no way of knowing if my answer is correct or not.
Can anybody help me with this please?
algebra-precalculus
edited Jan 12 at 7:59
Andrei
13.7k21230
asked Jan 12 at 7:37
user633979user633979
32
$endgroup$
add a comment |
$begingroup$
I am self-studying precalculus on my own. This is a problem from Ron Larson's precalculus 10e
The arch support of a bridge is
modeled by
$y = −0.0012x^2+
300$, where x and y are
measured in feet and the x-axis represents the ground.
(a) Use a graphing utility to graph the equation.
(b) Find one x-intercept of the graph. Explain how to
use the intercept and the symmetry of the graph to
find the width of the arch support.
For the first question (a) I graphed it on desmos
And for the second problem my approach was to use the distance formula and the answer was 1000feet. Since I am doing everything all on my own I have no way of knowing if my answer is correct or not.
Can anybody help me with this please?
algebra-precalculus
edited Jan 12 at 7:59
Andrei
13.7k21230
asked Jan 12 at 7:37
user633979user633979
32
$endgroup$
add a comment |
$begingroup$
I am self-studying precalculus on my own. This is a problem from Ron Larson's precalculus 10e
The arch support of a bridge is
modeled by
$y = −0.0012x^2+
300$, where x and y are
measured in feet and the x-axis represents the ground.
(a) Use a graphing utility to graph the equation.
(b) Find one x-intercept of the graph. Explain how to
use the intercept and the symmetry of the graph to
find the width of the arch support.
For the first question (a) I graphed it on desmos
And for the second problem my approach was to use the distance formula and the answer was 1000feet. Since I am doing everything all on my own I have no way of knowing if my answer is correct or not.
Can anybody help me with this please?
algebra-precalculus
edited Jan 12 at 7:59
Andrei
13.7k21230
asked Jan 12 at 7:37
user633979user633979
32
$endgroup$
I am self-studying precalculus on my own. This is a problem from Ron Larson's precalculus 10e
The arch support of a bridge is
modeled by
$y = −0.0012x^2+
300$, where x and y are
measured in feet and the x-axis represents the ground.
(a) Use a graphing utility to graph the equation.
(b) Find one x-intercept of the graph. Explain how to
use the intercept and the symmetry of the graph to
find the width of the arch support.
For the first question (a) I graphed it on desmos
And for the second problem my approach was to use the distance formula and the answer was 1000feet. Since I am doing everything all on my own I have no way of knowing if my answer is correct or not.
Can anybody help me with this please?
algebra-precalculus
algebra-precalculus
edited Jan 12 at 7:59
Andrei
13.7k21230
asked Jan 12 at 7:37
user633979user633979
32
edited Jan 12 at 7:59
Andrei
13.7k21230
asked Jan 12 at 7:37
user633979user633979
32
edited Jan 12 at 7:59
Andrei
13.7k21230
edited Jan 12 at 7:59
Andrei
13.7k21230
edited Jan 12 at 7:59
Andrei
13.7k21230
13.7k21230
asked Jan 12 at 7:37
user633979user633979
32
asked Jan 12 at 7:37
user633979user633979
32
asked Jan 12 at 7:37
user633979user633979
32
32
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Your answer is correct. Here is an example of how I would solve the problem.
We begin by finding the $x$-intercepts, using the fact that the $x$-intercepts are the points where $y = 0$
begin{align}
-0.0012x^2 + 300 = y &= 0 \
0.0012x^2 &= 300 \
x^2 = frac{300}{0.0012} &= 250000
end{align}
One solution to this equation is $x = sqrt{250000} = 500$, and since we can see that the graph is symmetric around the $y$-axis, we must have that $x = -sqrt{250000} = -500$ is also a solution. So we get the width of the arch support to be the difference $500 - (-500) = 1000$ feet between these two points, just as you got.
answered Jan 12 at 7:58
nesHannesHan
1785
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1 Answer
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1 Answer
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$begingroup$
Your answer is correct. Here is an example of how I would solve the problem.
We begin by finding the $x$-intercepts, using the fact that the $x$-intercepts are the points where $y = 0$
begin{align}
-0.0012x^2 + 300 = y &= 0 \
0.0012x^2 &= 300 \
x^2 = frac{300}{0.0012} &= 250000
end{align}
One solution to this equation is $x = sqrt{250000} = 500$, and since we can see that the graph is symmetric around the $y$-axis, we must have that $x = -sqrt{250000} = -500$ is also a solution. So we get the width of the arch support to be the difference $500 - (-500) = 1000$ feet between these two points, just as you got.
answered Jan 12 at 7:58
nesHannesHan
1785
$endgroup$
add a comment |
$begingroup$
Your answer is correct. Here is an example of how I would solve the problem.
We begin by finding the $x$-intercepts, using the fact that the $x$-intercepts are the points where $y = 0$
begin{align}
-0.0012x^2 + 300 = y &= 0 \
0.0012x^2 &= 300 \
x^2 = frac{300}{0.0012} &= 250000
end{align}
One solution to this equation is $x = sqrt{250000} = 500$, and since we can see that the graph is symmetric around the $y$-axis, we must have that $x = -sqrt{250000} = -500$ is also a solution. So we get the width of the arch support to be the difference $500 - (-500) = 1000$ feet between these two points, just as you got.
answered Jan 12 at 7:58
nesHannesHan
1785
$endgroup$
add a comment |
$begingroup$
Your answer is correct. Here is an example of how I would solve the problem.
We begin by finding the $x$-intercepts, using the fact that the $x$-intercepts are the points where $y = 0$
begin{align}
-0.0012x^2 + 300 = y &= 0 \
0.0012x^2 &= 300 \
x^2 = frac{300}{0.0012} &= 250000
end{align}
One solution to this equation is $x = sqrt{250000} = 500$, and since we can see that the graph is symmetric around the $y$-axis, we must have that $x = -sqrt{250000} = -500$ is also a solution. So we get the width of the arch support to be the difference $500 - (-500) = 1000$ feet between these two points, just as you got.
answered Jan 12 at 7:58
nesHannesHan
1785
$endgroup$
Your answer is correct. Here is an example of how I would solve the problem.
We begin by finding the $x$-intercepts, using the fact that the $x$-intercepts are the points where $y = 0$
begin{align}
-0.0012x^2 + 300 = y &= 0 \
0.0012x^2 &= 300 \
x^2 = frac{300}{0.0012} &= 250000
end{align}
One solution to this equation is $x = sqrt{250000} = 500$, and since we can see that the graph is symmetric around the $y$-axis, we must have that $x = -sqrt{250000} = -500$ is also a solution. So we get the width of the arch support to be the difference $500 - (-500) = 1000$ feet between these two points, just as you got.
answered Jan 12 at 7:58
nesHannesHan
1785
answered Jan 12 at 7:58
nesHannesHan
1785
answered Jan 12 at 7:58
nesHannesHan
1785
answered Jan 12 at 7:58
nesHannesHan
1785
1785
add a comment |
add a comment |
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