To find the vertical asymptotes, we must figure out when the denominator of \(\frac{5x^2-9}{3x^2+5x+2}\) equals 0. This will give us the vertical asymptotes. Let's do that!
\(3x^2+5x+2=0\) | I will use the AC method to factor out this. What number multiplies to get 6 and adds to get 5? 3 and 2, of course! Break up the b-term. | ||
\(3x^2+3x+2x+2=0\) | Let's solve this by grouping. I'll use parentheses to make it easier to follow, I hope. | ||
\((3x^2+3x)+(2x+2)=0\) | Factor out the GCF of both expressions inside of the parentheses. | ||
\(3x(x+1)+2(x+1)=0\) | Now, use the reverse-distributive-property to combine 3x and 2. In other words, \(ac+bc=(a+b)c\) | ||
\((3x+2)(x+1)=0\) | Set both factors equal to 0 and solve both independently. | ||
| Subtract the constant term from both sides of the equation. | ||
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x=-(2/3) and x=-1 are the location of both vertical asymptotes on the graph.
Now, add both answers together to get the result of \(a+b\).
\(-1-\frac{2}{3}=\frac{-3}{3}-\frac{2}{3}=-\frac{5}{3}=-1.\overline{66}\)
.Coincidentally, trigonometry was unnecessary in this problem (although use whatever solving method you'd like!).
The height of the rectangle (we know it is a rectangle because it a quadrilateral with 4 right angles) \(FLAX\) is of equal length to the height of\(\triangle ZAX\).
Let's draw in \(\overline{ZB}\) such that the segment is a perpendicular bisector of \(\overline{FL}\) and \(B\) lies on the midpoint of \(\overline{XA}\).
Inserting the altitude of an isosceles (or equilateral, in this case) triangle has 2 lesser-known properties.
1) The altitude bisects the intersected angle.
2) The altitude bisects the intersected segment.
Put all of this information together \(m\angle AZB=30^{\circ}\hspace{1mm}\text{and}\hspace{1mm}m\angle ZBA=90^{\circ}\hspace{1mm}\text{and}\hspace{1mm} m\angle BAZ=60^{\circ}\). \(AB=5\hspace{1mm}\text{and}\hspace{1mm} AZ=10\)
It's a 30-60-90 triangle. Yay! By definition, the ratio of the side lengths is \(1:\sqrt{3}:2\). Knowing this, we can calculate the length of the altitude.
\(\frac{ZB}{\sqrt{3}}=\frac{5}{1}\) | Multiply by the square root of 3 on both sides of the equation. |
\(ZB=5\sqrt{3}\) | |
It has already been established that ZB is the height of the rectangle. Therefore, multiply the length and the width together to get the total area.
\(A=10*5\sqrt{3}=50\sqrt{3}units^2\)
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