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Sagot :
Given: The height y of a ball (in feet) is given by the function y=-1/12x^2+2x+4 and x is the horizontal distance traveled by the ball.
Part A: How high is the ball when it leaves the child's hand?
Right after the ball leaves the child's hand, it has travelled 0 feet horizontally. Horizontal distance is represented by x, so we could say that x = 0.
Plug in 0 for our equation and solve for y, the height.
[tex]y=-\frac{1}{12}x^2+2x+4\\\\y=\frac{1}{12}\cdot0^2+2\cdot0+4\\\\y=0+0+4\\\\\boxed{y=4}[/tex]
Part B & C: How high is the ball at its maximum height?
What we basically want to do is find the vertex of the function.
There are multiple ways to do this. You could graph it or make a table, but this method is not efficient.
The method I am going to go over right now is putting the equation in vertex form.
[tex]y=-\frac{1}{12}x^2+2x+4[/tex]
Move the constant to the left side.
[tex]y-4=-\frac{1}{12}x^2+2x[/tex]
Factor out the x² coefficient.
[tex]y-4=-\frac{1}{12}(x^2-24x)[/tex]
Find out which number to add to create a perfect square trinomial.
(Half of 24 is 12, 12 squared is 144. We have to add 144/-12 (which is -12) to each side so that we end up with 144 inside the parentheses on the right side)
[tex]y-4-12=-\frac{1}{12}(x^2-24x+144)[/tex]
Factor the perfect square trinomial and simplify the right side.
[tex]y-16=-\frac{1}{12}(x-12)^2[/tex]
Isolate y on the left side.
[tex]y=-\frac{1}{12}(x+12)^2+16[/tex]
And now we are in vertex form.
Vertex form is defined as y = a(x-h)² + k with vertex (h, k).
In this case, our vertex is (12, 16).
You could've also taken the shortcut that for any quadratic f(x) = ax² + bx + c, the vertex (h, k) is (-b/2a, f(h)). That's basically a summation of this method which you can use if your teacher has taught it to you.
Part D & E: What is the horizontal distance travelled by the ball when it hits the ground?
When the ball hits the ground, y is going to be 0, since y is the ball's height.
There are many ways to solve a quadratic...split the middle, complete the square, and the quadratic formula.
[tex]-\frac{1}{12}x^2+2x+4=0[/tex]
Solving by splitting the midlde
If your quadratic has fractions, this is not a good option.
Solving by completing the square
Move the constant over the right side.
[tex]y=-\frac{1}{12}x^2+2x=-4[/tex]
Divide by the x² coefficient.
(Dividing by -1/12 is the same as multiplying by its reciprocal, -12.)
[tex]x^2-24x=-4\times-12[/tex]
Simplify the right side.
[tex]x^2-24x=48[/tex]
Halve the x coefficient, square it, and then add it to each side.
(Half of -24 is -12, and -12 squared is 144.)
[tex]x^2-24x+144=192[/tex]
Factor the perfect square trinomial.
[tex](x-12)^2=192[/tex]
Take the square root of each side.
[tex]x-12=\pm\sqrt{192}[/tex]
192 = 8 × 8 × 3, so we can simplify √192 to 8√3.
Add 12 to each side and we get our answer.
[tex]x=12\pm8\sqrt{3}[/tex]
Our function does not apply when x or y is less than 0, of course.
12-8√3 is negative, so this cannot be our answer.
So, the ball had travelled 12+8√3 feet at the time when it hit the ground.
Solving with the quadratic formula
For any equation ax² + bx + c = 0, the solution for x is [tex]\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex].
Our equation, y=-1/12x^2+2x+4, has a = -1/12, b=2, and c=4.
Let's plug these values into the quadratic formula.
[tex]\frac{-2\pm\sqrt{2^2-4\cdot\frac{-1}{12}\cdot4}}{2\cdot\frac{-1}{12}}=\frac{-2\pm\sqrt{4-\frac{-4}3}}{\frac{-1}6}=\frac{-2\pm\sqrt{\frac{16}{3}}}{\frac{-1}6}=\frac{-2\pm\frac{4}{\sqrt{3}}}{\frac{-1}6}[/tex]
Dividing by a fraction is the same as multiplying by its reciprocal...
[tex]-6(-2\pm\frac{4}{\sqrt{3}})=12\pm\frac{-24}{\sqrt{3}}=12\pm\frac{24}{\sqrt{3}}=12\pm\frac{24\sqrt{3}}3=\boxed{12\pm8\sqrt{3}}[/tex]
Of course, we only want the positive value, 12+8√3.
Revisiting Part B & C:
Since parabolae are symmetrical, if you know two values of x for some value of y (like the x-intercepts we just found in part B) then you can find the average between them to find what the x value of the vertex is, then plug that in to find the y value of the vertex (the height we want)
The average between 12+8√3 and 12-8√3 is 12. Plug that in and we get 16!
Part A: How high is the ball when it leaves the child's hand?
Right after the ball leaves the child's hand, it has travelled 0 feet horizontally. Horizontal distance is represented by x, so we could say that x = 0.
Plug in 0 for our equation and solve for y, the height.
[tex]y=-\frac{1}{12}x^2+2x+4\\\\y=\frac{1}{12}\cdot0^2+2\cdot0+4\\\\y=0+0+4\\\\\boxed{y=4}[/tex]
Part B & C: How high is the ball at its maximum height?
What we basically want to do is find the vertex of the function.
There are multiple ways to do this. You could graph it or make a table, but this method is not efficient.
The method I am going to go over right now is putting the equation in vertex form.
[tex]y=-\frac{1}{12}x^2+2x+4[/tex]
Move the constant to the left side.
[tex]y-4=-\frac{1}{12}x^2+2x[/tex]
Factor out the x² coefficient.
[tex]y-4=-\frac{1}{12}(x^2-24x)[/tex]
Find out which number to add to create a perfect square trinomial.
(Half of 24 is 12, 12 squared is 144. We have to add 144/-12 (which is -12) to each side so that we end up with 144 inside the parentheses on the right side)
[tex]y-4-12=-\frac{1}{12}(x^2-24x+144)[/tex]
Factor the perfect square trinomial and simplify the right side.
[tex]y-16=-\frac{1}{12}(x-12)^2[/tex]
Isolate y on the left side.
[tex]y=-\frac{1}{12}(x+12)^2+16[/tex]
And now we are in vertex form.
Vertex form is defined as y = a(x-h)² + k with vertex (h, k).
In this case, our vertex is (12, 16).
You could've also taken the shortcut that for any quadratic f(x) = ax² + bx + c, the vertex (h, k) is (-b/2a, f(h)). That's basically a summation of this method which you can use if your teacher has taught it to you.
Part D & E: What is the horizontal distance travelled by the ball when it hits the ground?
When the ball hits the ground, y is going to be 0, since y is the ball's height.
There are many ways to solve a quadratic...split the middle, complete the square, and the quadratic formula.
[tex]-\frac{1}{12}x^2+2x+4=0[/tex]
Solving by splitting the midlde
If your quadratic has fractions, this is not a good option.
Solving by completing the square
Move the constant over the right side.
[tex]y=-\frac{1}{12}x^2+2x=-4[/tex]
Divide by the x² coefficient.
(Dividing by -1/12 is the same as multiplying by its reciprocal, -12.)
[tex]x^2-24x=-4\times-12[/tex]
Simplify the right side.
[tex]x^2-24x=48[/tex]
Halve the x coefficient, square it, and then add it to each side.
(Half of -24 is -12, and -12 squared is 144.)
[tex]x^2-24x+144=192[/tex]
Factor the perfect square trinomial.
[tex](x-12)^2=192[/tex]
Take the square root of each side.
[tex]x-12=\pm\sqrt{192}[/tex]
192 = 8 × 8 × 3, so we can simplify √192 to 8√3.
Add 12 to each side and we get our answer.
[tex]x=12\pm8\sqrt{3}[/tex]
Our function does not apply when x or y is less than 0, of course.
12-8√3 is negative, so this cannot be our answer.
So, the ball had travelled 12+8√3 feet at the time when it hit the ground.
Solving with the quadratic formula
For any equation ax² + bx + c = 0, the solution for x is [tex]\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex].
Our equation, y=-1/12x^2+2x+4, has a = -1/12, b=2, and c=4.
Let's plug these values into the quadratic formula.
[tex]\frac{-2\pm\sqrt{2^2-4\cdot\frac{-1}{12}\cdot4}}{2\cdot\frac{-1}{12}}=\frac{-2\pm\sqrt{4-\frac{-4}3}}{\frac{-1}6}=\frac{-2\pm\sqrt{\frac{16}{3}}}{\frac{-1}6}=\frac{-2\pm\frac{4}{\sqrt{3}}}{\frac{-1}6}[/tex]
Dividing by a fraction is the same as multiplying by its reciprocal...
[tex]-6(-2\pm\frac{4}{\sqrt{3}})=12\pm\frac{-24}{\sqrt{3}}=12\pm\frac{24}{\sqrt{3}}=12\pm\frac{24\sqrt{3}}3=\boxed{12\pm8\sqrt{3}}[/tex]
Of course, we only want the positive value, 12+8√3.
Revisiting Part B & C:
Since parabolae are symmetrical, if you know two values of x for some value of y (like the x-intercepts we just found in part B) then you can find the average between them to find what the x value of the vertex is, then plug that in to find the y value of the vertex (the height we want)
The average between 12+8√3 and 12-8√3 is 12. Plug that in and we get 16!
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