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2 years ago

The graph of linear function g passes through the points (-7, -4) and (7, 6)

Mathematics

1 answer:

goldfiish [28.3K]2 years ago

7 0

Answer:

<h3>C. The slope is 5/7, and the y-intercept is 1.</h3>

Step-by-step explanation:

You must solve the slope-intercept form with the graph of the linear function g passing through the points.

Use the slope-intercept form.

$\underline{\text{SLOPE-INTERCEPT FORM:}}$

$\Longrightarrow: \sf{y=mx+b}$

M= slope
B= y-intercept.

Use the slope formula.

$\underline{\text{SLOPE:}}$

$\Longrightarrow \sf{\dfrac{y_2-y_1}{x_2-x_1} }$

(-7,-4) and (7,6)

y2=6
y1=(-4)
x2=7
x1=(-7)

$\sf{\dfrac{6-\left(-4\right)}{7-\left(-7\right)}}$

Solve.

$\sf{\dfrac{6-\left(-4\right)}{7-\left(-7\right)}=\sf{\dfrac{10}{14}=\dfrac{10\div2}{14\div2}=\dfrac{5}{7} }}$

Therefore, the slope is 5/7 and y-intercept is 1.
The correct answer is C. "The slope is 5/7, and the y-intercept is 1."

I hope this helps. Let me know if you have any questions.

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Consider the graph of the linear function h(x) = –x + 5. Which could you change to move the graph down 3 units?

Feliz [49]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
% left side templates
\begin{array}{llll}
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}\\\\
--------------------\\\\$

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

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

with that template in mind, let's see

$\bf h(x)=-x+5\implies h(x)=\stackrel{A}{-1}(\stackrel{B}{1}x\stackrel{C}{+0})\stackrel{D}{+5}$

a shift down by 3 units, means a vertical shift downwards, so D needs to drop by 3 units.

$\bf h(x)=\stackrel{A}{-1}(\stackrel{B}{1}x\stackrel{C}{+0})\boxed{\stackrel{D}{+5-3}}\implies h(x)=-1(1x+0)+2
\\\\\\
h(x)=-x+2$

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