Determine both the x and y intercepts. Please help! Will mark brainliest!

Mathematics

2 answers:

Mademuasel [1]3 years ago

3 0

Answer:

y-intercept: (0, 2.5)

x-intercept: (1.67, 0)

Step-by-step explanation:

Right now, you have 3x + 2y = 5. You want to change it into slope-intercept form to make the problem easier to solve.

2y = -3x + 5

y = -3/2x + 5/2

From here, you can see that the y-intercept is (0, 5/2). You can also say that the y-intercept is (0, 2.5).

To get the x-intercept, substitute 0 for y.

0 = -3/2x + 5/2

3/2x = 5/2

3x = 5

x = 5/3

So, the x-intercept is (5/3, 0). You can also say that the x-intercept is (1.67, 0).

Hope this helps!

Vitek1552 [10]3 years ago

3 0

Answer:

Step-by-step explanation:

To find the x-intercept, let y = 0. Then 3x = 5 and x = 5/3, or x = 1.667, and the x-intercept is (1.667, 0).

To find the y-intercept, let x = 0. Then 2y = 5 and y = 5/2 or 2.5. Then the y-intercept is (0, 2.5).

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$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\

\begin{array}{rllll} 
% left side templates
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}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\end{array}$

$\bf \begin{array}{llll}


\bullet \textit{ vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}
\end{array}$

now, with that template above in mind, let's see this one

$\bf parent\implies f(x)=|x|
\\\\\\
\begin{array}{lllcclll}
f(x)=&3|&1x&+2|&+4\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}$

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C=2, horizontal shift by C/B or 2/1 or just 2, to the left
D=4, vertical shift upwards of 4 units

check the picture below