PAM - Systems of Equations Using Linear and Quadratic Equations Lesson
Systems of Equations Using Linear and Quadratic Equations
In this section we will revisit linear and quadratic equations as a combined topic. What do they allow us to discover concerning new topics? Intersections!
Earlier, in Heart of Algebra module, we studied lines and discovered how to find the intersection of lines. Here we will use the techniques we have learned to find the intersection of lines and quadratic equations. Remember quadratic equations are of the general form y= ax² + bx + c, but may also be in vertex form.
Parabola and Horizontal Line Intersection
From our earlier study of horizontal lines, we know that these are parallel to the x-axis. It is important to remember why their slope is zero. The slope of the horizontal line is zero because the numerator, the difference of the y-values between two points is zero. Thus division by anything into zero has an answer of zero.
Example 1: The intersection of a horizontal line and a quadratic (parabolic) equation.
Let's look at the graph below of the equations:
y = 4
y = x²
Now let's analyze the graph above.
The linear equation is y = 4, a horizontal equation with a slope of zero.
The quadratic equation y = x², is concave up, and has its vertex at (0, 0), a minimum.
Note that the two equations intersect. From the graph it is easy in this case to see that they intersect two times, at points (-2, 4) and (2, 4). But what if we don't have a graph? What techniques can we use to solve the system of equations?
Remember when we solved one equation for a variable and then substituted our solution in the other linear equation to solve. We will use the technique of substitution to locate the intersections.
y = x²
4 = x² Substitute using the linear equation y = 4, substitute 4 for y
Solve for x by taking the square root of both sides
±2 = x Remember a square root yields two answers (check for validity of both based on the problem context). Both are valid in the context of finding the intersection of the equations.
Knowing that the x-values of the intersections are x = −2 and x = 2. Substitute both values into the equation y = x² to find the y-values and create your points.
y = (-2)² and y = (2)²
y = 4 y = 4
The horizontal line equation.
(-2, 4) (2, 4) Points of intersection.
Always a good idea to put the substituted value in ( ) so that any negative sign is considered correctly.
Note the comment above concerning the horizontal line equation y = 4. Did we have to calculate and substitute to find y for the horizontal line intersection? No, as we already knew y. It is, however, a good check to make sure that we have found the correct x-values for the intersection.
Reasoning Questions - Set 1
Parabola and Vertical Line Intersection
From the Heart of Algebra module, remember that vertical lines are up and down on the page. They also have no slope as when the slope formula is used, the denominator becomes zero, and division by zero is illegal.
Example 2: The intersection of a vertical line and a quadratic equation. Let's look at the two equations below
x = -2
y = x² - 3x - 4
Notice that the vertical line x = −2 crosses the parabola in one point. This is great because it means that we already know the x-value. We only have to find the y-value to have the point ordered pair. So substitute in x = −2 and solve for y.
f(−2) = (−2)² - 3(−2) - 4 Substitute x = −2. Note the use of function notation to document the substitution y = f(x).
y = f(−2) = 4 + 6 - 4 = 6 The point is found and documented easily for remembering when function notation is used.
(−2, 6) The intersection point.
Important Note: Look at the graph. What is the range increments on the y-axis graph? The first major block is 25 with 4 gridlines prior, so the scale for the gridlines is = 6.25. Thus the scale for the y-axis is 6.25, not 6, although the intersection point appears to be on the gridline.
What is the scale of the x-axis? The first number shown is 4, on a major block with four gridlines before it. The scale of the x-axis is 
= 1, or 1 small block equals one unit.
Now you can understand why it is good to check the scale of the x-axis and y-axis. It is just as important here as when we use multiplication to enlarge or shrink the sides of objects by a positive or negative increment or create dilations using multiplication.
Reasoning Questions - Set 2
Parabola and Slant Line Intersection
Example 3: The intersection of a quadratic equation (parabola) and a slant line.
What is a slant line? A slant line is a line that does not have slope of zero or undefined. The slant line could have a positive slope, m > 0, or a negative slope, m < 0.
Let's examine this concept using the equations and graph below.
The equation of the line is
y=−x+1 and the equation of the parabola is
y=x2+2x+1
-x + 1 = x² + 2x + 1 Substitute (-x + 1) for y since the two equations must have the same x and y when they intersect.
0 = x² + 3x Combine like terms setting the variables equal to zero. Add x to both sides and subtract 1 from both sides.
0 = x(x + 3) Factor to solve for x. Take out x as the greatest common factor.
x = 0 and x + 3 = 0 Set each factor = 0 as either one or the other or both must be zero in order for the multiplied factors to equal zero.
x = 0 and x = -3 Solve for x.
Solve for y to find the points of intersection. Use either equation, but the easiest one is the best.
y = -0 + 1 = 1 Substitute in x = 0 and solve for y.
y = -(-3) + 1 = 4 Substitute in x = -3 and solve for y.
Intersection points are (0, 1) and (-3, 4) and we can verify by examining the graph. These can be checked by substituting in the opposite equation.
Note: If the equation formed is not easily factorable, using the quadratic formula will provide the intersection points by finding the possible x-values.
Watch the following video for another look at finding intersections of lines and parabolas. This video will explore what must happen if substitution does not lead to easy factoring.
Reasoning Questions - Set 3
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