Advanced Examples

This is another FREE ALGEBRA PRINTABLE presented to you from the Algebra section of K12math.com

1) Plot: Complete the square to get standard form.    (x2 - 6x + 9)  +   (y2 + 4y + 4)    +  4 - 9 - 4 = 0 (x2 - 6x + 9) + (y2 + 4y + 4) = 9 (x - 3)2 + (y + 2)2 = 32 This is a circle whose origin is (3,-2) and whose radius is 3. 2) Plot: Complete the square for the x terms to get x2 - 10x + 25    - 32y - 71    - 25 = 0 (x - 5)2 = 32y + 96 (x - 5)2 = 32( y + 3 ) 4c = 32, c = 8 This is a parabola whose vertex is at (5, -3) , and it opens along the positive y axis. 3) Plot: Complete the square for both x and y terms: first factor out the 2 from the x terms 2( x2 + 2x + 1) + (y2 +2y + 1)   - 5    - 3   = 0 (We add -3 since the x terms required +2 and the y terms required +1, 2+1 = 3 so we must subtract this 3 as well.) 2(x + 1)2 + (y + 1)2 - 8 = 0 2(x + 1)2 + (y + 1)2 = 8 Now divide both sides by 8   This is an ellipse with a = 2, b = 2.8 and its center is at (-1, -1).   b > a so it major axis is along the Y axis. foci: (-1, (-1 ±c)) (-1, -3) and (-1, 1) 4) Plot: Complete the squares: x2 - 8x +16 - 4 (y2 - 16y + 64 ) - 256 - 16 + 256 = 0 (x - 4 )2 - 4 (y - 8)2 = 16 Now divide by 16   This is a hyperbola with a = 4, b = 2 and center at (4, 8). 5) Where does this line intersect this parabola? Find exact values and show graphically.              y = 5 - 2x              x2 = 4y Substitute (5 - 2x) for y in the equation for the parabola and solve for x. x2 = 4(5 - 2x) x2 = 20 - 8x x2 + 8x - 20 = 0 (x + 10)(x - 2) = 0 x = 2, -10 y = 5 - 2(2) = 1 y = 5 - 2(-10) = 25 The line intersects the parabola at two points: (2,1) and (-10,25) 6) Where do these two parabolas intersect? x2 = 4y x2 - 6x +4y - 40 = 0 We need to get the second equation in standard form by completing the square: x2 - 6x + 9 + 4y - 40 - 9 = 0 (x - 3)2 + 4y - 40 = 0 (x - 3)2 = -4y + 40 (x - 3)2 = -4(y - 10) This is a parabola whose vertex is ( 3, 10) and it opens down the negative Y axis. Substituting x2/4 for y in the second equation we get: x2 - 6x +4(x2/4) - 40 = 0 2x2 -6x - 40 = 0 Using the quadratic formula with a=2, b = -6 and c = -40 we get x = -3.2 and 6.2 Substitute these back into the first equation to get y: (-3.2)2/4 = 2.6 and (6.2)2/4 = 9.6 for y. The points of intersection are: (-3.2, 2.6) and (6.2,9.6) 7) Where does this hyperbola and ellipse intersect? Clear the fractions by multiplying the first equation by 16 and the second equation by 36. x2 - 4y2 = 16 x2 + 9y2 = 36 Now subtract the second equation from the first equation. -4y2 -9y2 = 16 - 36 simplify -13y2 = -20 y2 = -20/-13 Use the first equation to solve for x. The hyperbola and ellipse intersect at four points shown above, which are: (4.71, 1.24) (4.71, -1.24) (-4.71, 1.24) (-4.71, -1.24) 8) Where does the parabola (x + 4)2 = 32(y + 2) a) cross the Y axis? x is 0 for the y intercept, substituting in we have: 42 = 32(y + 2) 16 = 32(y + 2) 1/2 = y + 2 -3/2 = y The y intercept is -1.5 b) cross the X axis? y is 0 for the x intercepts, substituting in we have: (x + 4)2 = 32(2) = 64 Now take the square root of both sides. x + 4 = ±8 x = 8 - 4 and x = -8 - 4 The x intercepts are 4 and -12 9) Where does the parabola x2 + 32x + 8y + 280 = 0 cross the X axis? To answer this question we need this equation in standard form. Complete the square for x. x2 + 32x +256 + 8y + 280 - 256 = 0 (x +16)2 + 8y + 24 = 0 (x +16)2 = -8y - 24 (x + 16)2 = -8(y + 3) Looking at the equation we see that the vertex (h,k) = (-16, -3 ) and it is below the X axis ( k is negative). c = -8/4 = -2 which means the parabola opens along the negative Y axis. Taken together this parabola cannot cross the X axis. Algebraically we set y = 0 and solve for x (in the original equation we get): x2 - 32x + 280 = 0 checking the discriminant in the quadratic formula we have (-32)2 - 4 (1)(280) = -96 which means there are no real solutions for x and, therefore, this parabola cannot cross the X axis. 10) Plot the equation 3x2 - 12x + 2y2 + 20y +10 = 0 First we need to write this equation in standard form. To do this we complete the square. Before we begin we need to recall that the standard equations do not have coefficients for x and y in the binomial terms: for example, for a circle: (x - h)2 + (y - k)2 = 1 (x - h)2 = x2 -2hx +h2 Our equation starts off with 3x2 not x2. There's a 2y2 but we need just y2. The technique to use is to factor the coefficient for x2 out of the x terms and factor out the coefficient for y2 out of the y terms. Then complete the square. 3x2 - 12x + 2y2 + 20y +10 = 0   First factor: 3( x2 - 4x ) + 2( y2 +10y ) +10 = 0 In the first line we need a +4 with the x terms and in the second line we need a +25 with the y terms. 3( x2 - 4x + 4 ) + 2( y2 +10y + 25 ) +10 = 0 Here's the critical point, we need to determine how to adjust the equation for our additional constants. Multiply it back out: 3( x2 - 4x + 4 ) = 3x2 - 12x + 12 and 2( y2 +10y + 25 ) = 2y2 +20y + 50 Altogether we've modified the equation by 12 + 50 = 62 so we need to subtract 62. Here it is: 3( x2 - 4x + 4 ) + 2( y2 +10y + 25 ) + 10 - 62 = 0 3(x - 2)2 + 2(y +5)2 = 52 52, 3, and 2 now must be come ones. Divide the equation by 52: We're almost there. note: So: Our equation becomes: In this case decimals in the denominator are ok. We have an ellipse centered at (2, -5) and its major axis is along the Y axis (since 26 > 17.3) Take a minute here. Recall that simplifying a fraction with a denominator that is a fraction requires you to first invert the fraction in the denominator then multiply the numerator by it. The same applies in reverse. A fraction in the numerator can be inverted then moved to the denominator of the original fraction multiplying that denominator. 11) graphically determine if the following conics intersect. The first conic is an ellipse: center is (-1,-10) major axis is along the X axis. The second conic is a hyperbola: center is the origin branches open along the positive and negative X axis. The Y range for the ellipse is -10 ± 2 that is -12 to -8 and the X range for the ellipse is -1 ± 4 that is -5 to 3. The vertices for the hyperbola are at (-6, 0) and (6,0). The X range for this hyperbola is therefore x <= -6 or x >= 6 which is outside the X range for the ellipse so they do not intersect. In general, when conics are not centered at the origin and you are tasked with determining whether or not they intersect, algebraically doing this results in an equation of degree 4. It is easier to graph them to see if there are any intersections. Example 7 allowed an algebraic solution since both conics were centered at the origin. If the conics share the same center then you can translate them back to the origin, solve for the points of intersection, then translate these coordinates back. The following example illustrates this technique. 12) Find the intersections of these two conics: (x-4)2 - 4(y+9)2 = 16 (x-4)2 + 9(y+9)2 = 36 The center for these conics is the same, (4, -9). Conceptually we translate the coordinate system for these equations to a new coordinate system +4 units along X and -9 units along Y. Using this new coordinate system we can eliminate the -4 and +9 from the equations. The original coordinate system is shown with the dashed lines. The new coordinate system is shown in green. And for each system each tic mark represents 2 units. Using this coordinate system our equations become x2 - 4y2 = 16 x2 + 9y2 = 36 The solution to these equations were found in example 7 to be: (4.71, 1.24) (4.71, -1.24) (-4.71, 1.24) (-4.71, -1.24) We realize that these are coordinates in the green coordinate system. We need to translate them back to the original coordinate system by adding 4 to each x value and subtracting 9 from each y value to get the solution: (8.71, -8.24) (8.71, -10.24) (4.71, -8.24) (4.71, -10.24)

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