How do you do systems of two equations? Problem: y=-3x+4, y=3x-2

1 answer:

8 0

I assume you are trying to solve for each variable. if thats the case you want to use one equation to solve for y and then plug the y value into the next variable
Problem: y =-3x=4 so -3x+4=3x-2
x=1
plug x value into one of the equations
y=3(1)-2
y=1

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The graph of F(x) shown below has the same shape as the graph of G(x) = x^2 but it is shifted down 5 units and to the left 4 uni

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$\bf ~\hspace{10em}\textit{function transformations} \\\\\\ \begin{array}{llll} f(x)= A( Bx+ C)^2+ D \\\\ f(x)= A\sqrt{ Bx+ C}+ D \\\\ f(x)= A(\mathbb{R})^{ Bx+ C}+ D \end{array}\qquad \qquad \begin{array}{llll} f(x)=\cfrac{1}{A(Bx+C)}+D \\\\\\ f(x)= A sin\left( B x+ C \right)+ D \end{array} \\\\[-0.35em] ~\dotfill\\\\ \bullet \textit{ stretches or shrinks horizontally by } A\cdot B\\\\ \bullet \textit{ flips it upside-down if } A\textit{ is negative}\\ ~~~~~~\textit{reflection over the x-axis}$

$\bf \bullet \textit{ flips it sideways if } B\textit{ is negative}\\ ~~~~~~\textit{reflection over the y-axis} \\\\ \bullet \textit{ horizontal shift by }\frac{ C}{ B}\\ ~~~~~~if\ \frac{ C}{ B}\textit{ is negative, to the right}\\\\ ~~~~~~if\ \frac{ C}{ B}\textit{ is positive, to the left}\\\\ \bullet \textit{ vertical shift by } D\\ ~~~~~~if\ D\textit{ is negative, downwards}\\\\ ~~~~~~if\ D\textit{ is positive, upwards}\\\\ \bullet \textit{ period of }\frac{2\pi }{ B}$

with that template in mind, let's see

down by 5 units, D = -5

to the left by 4 units, C = +4

$\bf G(x)=x^2\implies G(x)=1(1x+\stackrel{C}{0})^2+\stackrel{D}{0} \\\\\\ \begin{cases} D=-5\\ C=+4 \end{cases}\implies F(x)=1(1x+\stackrel{C}{4})^2\stackrel{D}{-5}\implies F(x)=(x+4)^2-5$