Y=-2x+2 y = 7x +11 Solve systems by substitution

Mathematics

1 answer:

Agata [3.3K]2 years ago

5 0

Answer: (-1,4)

Step-by-step explanation:

y=-2x+2 y=7x+11 <----- substitute this equation for y in the

equation y=-2x+2

7x+11=-2x+2 Add 2x to both sides

9x+11=2 Subtract 11 from both sides

9x=-9 Divide both sides by 9

x=-1

y=-2(-1)+2 Plug in x to one of the equations, I chose y=-2x+2

y=2+2 because it was the easiest, then solve

y=4

Solution is (-1,4)

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Suppose f (x) = x^2. Find the graph of f(x+2)

NISA [10]

$\bf ~~~~~~~~~~~~\textit{function transformations}
\\\\\\
% templates
f(x)={{ A}}({{ B}}x+{{ C}})+{{ D}}
\\\\
~~~~y={{ A}}({{ B}}x+{{ C}})+{{ D}}
\\\\
f(x)={{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\\\
f(x)={{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\\\
f(x)={{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\\\\
--------------------$

$\bf \bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}\\
~~~~~~\textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }{{ B}}\textit{ is negative}\\
~~~~~~\textit{reflection over the y-axis}$

$\bf \bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
~~~~~~if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by }{{ D}}\\
~~~~~~if\ {{ D}}\textit{ is negative, downwards}\\\\
~~~~~~if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}$

with that template in mind,

$\bf f(x)=x^2\qquad \quad f(x+2)=(x+2)^2\implies f(x+2)=\stackrel{A}{1}(\stackrel{B}{1}x\stackrel{C}{+2})^2\stackrel{D}{+0}$

C = 2 B = 1 C/B = 2/1 or +2, horizontal left shift of 2 units

f(x) shifted left by 2 units is f(x+2).

8 0

2 years ago

What’s the answer to this?

Elina [12.6K]

Step-by-step explanation:

Line is passing through the points:

$(-2,\:3)=(x_1,\:y_1) \:\&\: (2,\:5)=(x_2 ,\:y_2)$

Equation of line in two point form is given as:

$\frac{y -y_1 }{y_1 -y_2 } = \frac{x -x_1 }{x_1 -x_2 } \\ \\ \therefore \frac{y -3}{ 3 -5} = \frac{x -(-2) }{-2 -2} \\ \\ \therefore \frac{y -3}{ - 2} = \frac{x -(-2) }{-4} \\ \\ \therefore \frac{y - 3}{1} = \frac{x +2}{2} \\ \\ \therefore \: y-3= \frac{x}{2} + \frac{2}{2} \\ \\\therefore \: y= \frac{x}{2} + 1 +3\\ \\ \huge \red{ \boxed{\therefore \: y= \frac{1}{2} \:x+ 4}}\\ is \: the \: required \: equation \: of \: line.$