First, I know that we have 40 yards of fencing material. Our equation to finding out the perimeter of the the rectangle is 2W+2L=40. I also know that the formula for finding out the area is LW=A.The L variable stands for length and the W stands for width.
Second, to be able to solve this problem these were the steps that I followed. The first thing I did was I got L by itself. I subtracted 2W to both sides. Then I divided 2 on both sides. I found that L equals -W+20.
Next, to find out what the W was, I plugged in the L value of the perimeter into LW, which is the area formula. Then I distributed the W times the L value and I got -W2+20W. To find the x of my vertex, I plugged -W2+20W into -b/2a. I got 10, which was my W. I plugged my W value back into the perimeter formula, which was 2W+2L=40. I multiplied 2 and 10 and got 20. Then I …show more content…
First I set the equation W(-W+20) equal to 0. Then I broke it into 2 parts, W=0 and -W+20=0. For the first part the x-intercept was (0,0). I found the second x-intercept by solving -W+20=0. I subtracted -20 to both sides. My W came out to 20, which meant that my second x-intercept was (20,0). The next step I took was I found out my vertex. I already know the vertex because I used the vertex formula earlier. Lastly, I graphed my vertex and x-intercepts, which made a parabola.
In conclusion, by isolating the L and substituting the L value into the area formula, plugging it into the vertex formula, plugging it back to the perimeter formula, and multiplying the L and the W value together. I was able to found the vertex and x-intercepts. Which gave me the 3 point that allowed me to graph a