The right answer is -2/3
please see the attached picture for full solution
Hope it helps..
Good luck on your assignment..
We have the function
and we want to find a function that has the same y-intercept than the previous function.
First, let's find the y-intercept by subtituting 0 for 'x'.

Now that we found that y-intercept =-3, any lineal function of the type:
will have the same y-intercept. Where 'a' can take all the real values.
Also, any quadratic function of the type:
will have the same y-intercept. Where 'a' and 'b' can take all the real values.
Answer:
<em><u>given </u></em><em><u>:</u></em><em><u>-</u></em>
<em><u>for </u></em><em><u>rectangular</u></em><em><u> </u></em><em><u>part:</u></em><em><u> length</u></em><em><u>=</u></em><em><u>1</u></em><em><u>2</u></em><em><u>i</u></em><em><u>n</u></em><em><u>,</u></em><em><u> breadth</u></em><em><u>=</u></em><em><u>8</u></em><em><u>i</u></em><em><u>n</u></em>
<em><u>for</u></em><em><u> </u></em><em><u>triangular</u></em><em><u> </u></em><em><u>part:</u></em><em><u>base=</u></em><em><u>8</u></em><em><u>i</u></em><em><u>n</u></em><em><u>,</u></em><em><u> </u></em><em><u>height=</u></em><em><u>3</u></em><em><u>i</u></em><em><u>n</u></em>
<em><u>area of the given fig:</u></em>
<em><u>area of the given fig:area of 2 triangles +area of rectangle </u></em>
<em><u>
</u></em>
<h2>
<em><u>hope</u></em><em><u> it</u></em><em><u> helps</u></em><em><u> </u></em><em><u>you</u></em><em><u><</u></em><em><u>3</u></em></h2>
<h3 />
Don't touch the center. It is already even.
Start anywhere by connecting a dotted line from one vertex to the next. To keep things so we know what we are talking about, go clockwise. Now you have 2 points that are Eulerized that were not before.
Skip and edge and do the same thing to the next two vertices. Those two become eulerized. Skip an edge and do the last 2.
Let's try to describe this better. Start at any vertex and number them 1 to 6 clockwise.
Join 1 to 2
Join 3 to 4
Join 5 to 6
I think 3 is the minimum.
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