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laila [671]
1 year ago
9

In the relationship shown by the data linear ? If so , model the data with an equation A. The relationship is not linear B. The

relationship is linear; y+2=4/5 (x+9) C . The relationship is linear; y + 9 = - 4/5 (x+2) D. The relationship is linear; y+ 2 = -5/4 (x+9)

Mathematics
1 answer:
dybincka [34]1 year ago
7 0

Let's take two points so that we can get the equation of the line which goes through those points. P1 (-9, -2), P2 (3, -17):

m=\frac{y_2-y_1}{x_2-x_1}=\frac{-17-(-2)}{3-(-9)}=\frac{-17+2}{3+9}=-\frac{15}{12}=-\frac{5}{4}\begin{gathered} y-y_1=m(x-x_1) \\ y-(-2)=-\frac{5}{4}\cdot(x-(-9)) \\ y+2=-\frac{5}{4}\cdot(x+9)_{} \\ y=-\frac{5}{4}x-\frac{45}{9}-2 \\ y=-\frac{5}{4}x-7 \\ f(x)=-\frac{5}{4}x-7 \end{gathered}

So, y is the line which goes through the first and last points of the chart.

To proof that the rest of points go through the line as well, we will evalute each point

\begin{gathered} f(-5)=-\frac{5}{4}\cdot(-5)-7=\frac{25}{4}-7=-\frac{3}{4}\ne-7 \\ f(-1)=-\frac{5}{4}(-1)-7=\frac{5}{4}-7=-\frac{23}{4}\ne-12 \end{gathered}

Since the evaluation of these points don't correspond to the values of the chart we can assure that the relationship is not linear

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Given the function f(x);

f(x)=-4x^2+4x-1

Evaluating the function f(x+h);

\begin{gathered} f(x+h)=-4(x+h)^2+4(x+h)-1 \\ f(x+h)=-4(x^2+2xh+h^2)^{}+4(x+h)-1 \\ f(x+h)=-4x^2-4h^2-8xh^{}+4x+4h-1 \end{gathered}

So;

f(x+h)=-4x^2-4h^2-8xh^{}+4x+4h-1

Evaluating the second function;

\begin{gathered} \frac{f(x+h)-f(x)}{h}=\frac{-4x^2-4h^2-8xh^{}+4x+4h-1-(-4x^2+4x-1)}{h} \\ \frac{f(x+h)-f(x)}{h}=\frac{-4x^2-4h^2-8xh^{}+4x+4h-1+4x^2-4x+1}{h} \\ \frac{f(x+h)-f(x)}{h}=\frac{-4x^2+4x^2-4h^2-8xh^{}+4x-4x+4h-1+1}{h} \\ \frac{f(x+h)-f(x)}{h}=\frac{-4h^2-8xh^{}+4h}{h} \end{gathered}

simplifying further;

\begin{gathered} \frac{f(x+h)-f(x)}{h}=\frac{-4h^2-8xh^{}+4h}{h}=-4h-8x+4 \\ \frac{f(x+h)-f(x)}{h}=-4h-8x+4 \end{gathered}

Therefore, we have;

undefined

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