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2 years ago

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Which input was substituted into the function f(x) = x to create the given graph? f(x + 2.5) f(x – 2.5) f f (StartFraction 5 Ove

r 2 EndFraction x) f f (StartFraction 2 Over 5 EndFraction x)

1 answer:

amm18122 years ago

6 0

When a coordinate plane with a line graphed passing through (negative 2.5, negative 5) and (2.5, 0), the input that was substituted into the function f(x) = x to create the given graph is f(x-2.5).

<h3>What is a coordinate plane?</h3>

A coordinate plane is a two-dimensional plane formed by the intersection of two number lines.

In this case, when a coordinate plane with a line graphed passing through (negative 2.5, negative 5) and (2.5, 0), the input that was substituted into the function f(x) = x to create the given graph is f(x-2.5).

Learn more about coordinate plane on:

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The height y of a ball (in feet) is given by the function and x is the horizontal distance traveled by the ball.

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Given: <em>The height y of a ball (in feet) is given by the function </em><em>y=-1/12x^2+2x+4 </em><em>and x is the horizontal distance traveled by the ball.</em>

Part A:<em> </em><em>How high is the ball when it leaves the child's hand?</em>

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.

$y=-\frac{1}{12}x^2+2x+4\\\\y=\frac{1}{12}\cdot0^2+2\cdot0+4\\\\y=0+0+4\\\\\boxed{y=4}$

Part B & C: <em>How high is the ball at its maximum height?
</em>
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.

$y=-\frac{1}{12}x^2+2x+4$

Move the constant to the left side.

$y-4=-\frac{1}{12}x^2+2x$

Factor out the x² coefficient.

$y-4=-\frac{1}{12}(x^2-24x)$

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)

$y-4-12=-\frac{1}{12}(x^2-24x+144)$

Factor the perfect square trinomial and simplify the right side.

$y-16=-\frac{1}{12}(x-12)^2$

Isolate y on the left side.

$y=-\frac{1}{12}(x+12)^2+16$

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: <em>What is the horizontal distance travelled by the ball when it hits the ground?</em>
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.

$-\frac{1}{12}x^2+2x+4=0$
<u>
</u><u>Solving by splitting the midlde</u>
If your quadratic has fractions, this is not a good option.
<u>
</u><u>Solving by completing the square</u>
Move the constant over the right side.

$y=-\frac{1}{12}x^2+2x=-4$

Divide by the x² coefficient.
(Dividing by -1/12 is the same as multiplying by its reciprocal, -12.)

$x^2-24x=-4\times-12$

Simplify the right side.

$x^2-24x=48$

Halve the x coefficient, square it, and then add it to each side.
(Half of -24 is -12, and -12 squared is 144.)

$x^2-24x+144=192$

Factor the perfect square trinomial.

$(x-12)^2=192$

Take the square root of each side.

$x-12=\pm\sqrt{192}$

192 = 8 × 8 × 3, so we can simplify √192 to 8√3.
Add 12 to each side and we get our answer.

$x=12\pm8\sqrt{3}$

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.

<u>Solving with the quadratic formula</u>
For any equation ax² + bx + c = 0, the solution for x is $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ .

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.

$\frac{-2\pm\sqrt{2^2-4\cdot\frac{-1}{12}\cdot4}}{2\cdot\frac{-1}{12}}=\frac{-2\pm\sqrt{4-\frac{-4}3}}{\frac{-1}6}=\frac{-2\pm\sqrt{\frac{16}{3}}}{\frac{-1}6}=\frac{-2\pm\frac{4}{\sqrt{3}}}{\frac{-1}6}$

Dividing by a fraction is the same as multiplying by its reciprocal...

$-6(-2\pm\frac{4}{\sqrt{3}})=12\pm\frac{-24}{\sqrt{3}}=12\pm\frac{24}{\sqrt{3}}=12\pm\frac{24\sqrt{3}}3=\boxed{12\pm8\sqrt{3}}$

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!

5 0

3 years ago