F(x)=x^2-10 and g(x)=7–x, find (f-g)(-6)

1 answer:

3 0

$(f-g)(x)=f(x)-g(x)\\\\f(x)=x^2-10\\\\f(-6)\to\text{put x = -6 to the equation}\\\\f(-6)=(-6)^2-10=36-10=26\\\\g(x)=7-x\\\\g(-6)\to\text{put x = -6 to the equation}\\\\g(-6)=7-(-6)=7+6=13\\\\(f-g)(-6)=f(-6)-g(-6)=26-13=13$

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during a softball game kay hit a fly ball the function f(x) = -16t^2 + 64t + 4 describes the height of the softball in feet. mak

STatiana [176]

<h2>Hello!</h2>

The graph is attached.

To graph a parabola we need to know the following:

- If the parabola is open upwards or downwards

- They axis intercepts (if they exist)

- The vertex position (point)

We are given the function:

$f(t)=-16t^{2}+64t+4$

Where,

$a=-16\\b=64\\c=4$

For this case, the coefficient of the quadratic term (a) is negative, it means that the parabola opens downwards.

Finding the axis interception points:

Making the function equal to 0, we can find the x-axis (t) intercepts, but since the equation is a function of the time, we will only consider the positive values, so:

$f(t)=-16t^{2}+64t+4\\0=-16t^{2}+64t+4\\-16t^{2}+64t+4=0$

Using the quadratic equation:

$\frac{-b+-\sqrt{b^{2}-4ac } }{2a}=\frac{-64+-\sqrt{64^{2}-4*-16*4} }{2*-16}\\\\\frac{-64+-\sqrt{64^{2}-4*-16*4} }{2*-16}=\frac{-64+-\sqrt{4096+256} }{-32}\\\\\frac{-64+-\sqrt{4096+256} }{-32}=\frac{-64+-(65.96) }{-32}\\\\t1=\frac{-64+(65.96) }{-32}=-0.06\\\\t2=\frac{-64-(65.96) }{-32}=4.0615$

So, at t=4.0615 the height of the softball will be 0.

Since we will work only with positive values of "x", since we are working with a function of time:

Let's start from "t" equals to 0 to "t" equals to 4.0615.

So, evaluating we have:

$f(0)=-16(0)^{2}+64(0)+4=4\\\\f(1)=-16(1)^{2}+64(1)+4=52\\\\f(2)=-16(2)^{2}+64(2)+4=68\\\\f(3)=-16(3)^{2}+64(3)+4=52\\\\f(4.061)=-16(4.0615)^{2}+64(4.0615)+4=0.0034=0$