## Lesson 7: Solving Linear-Quadratics Systems

When given the equation of a linear function and a quadratic equation, you can find the point of intersection of the two systems. The two systems may never intersect, may intersect at one point, or at two points. When they intersect at one point, the line is a tangent line. When they intersect at two points, the line is called a secant. Before you try to find the points of intersection you may first wish to know if there are any points of intersection. You can use the discriminant to see if a system has one, two, or no points of intersection. The formula of the discriminant is b²-4ac. If the discriminant is less than zero, there are no solutions. If the discriminant is zero, there is only one solution. If the discriminant is greater than zero, there are two solutions.

Let's look at a sample linear quadratic system and see if it has no solution, one solution, or two solutions. Suppose we are given the following equations:

y=3x+5 and y=3x²-2x-4

First, equate the equations:

3x+5=3x²-2x-4

3x²-5x-9=0

Now, determine a, b, and c.

ax²+bx+c=0

3x²-5x-9=0

So, a=3, b=-5, and c=-9.

Now plug in the values of a, b, and c into the discriminant, b²-4ac.

b²-4ac

=(-5)²-4(3)(-9)

=25-12(-9)

=25+108

=133

Since the value of the discriminant is above zero, the line intersects the quadratic function at two points. Therefore, the line is a secant line. Now let us find the co-ordinates of the two points of intersection. To do this equate the two equations, and then simplify. Afterwards, find the values of x. Finally, plug in the values of x into either of the original equations to find y. Let's go through an example:

Find the co-ordinates of the two points of intersection for the following quadratic-linear system:

y=2x²-10x+15 and y=9x-15

2x²-10x+15=9x-15 First equate the two equations. (Make them equal).

2x²-19x+30=0 Simplify. Now you have a quadratic equation. Solve it.

2x²-4x-15x+30=0 In this case you can use factoring to solve the equation.

2x(x-2)-15(x-2)=0

(2x-15)(x-2)=0

x=7.5 or x=2

Now we must find the y-coordinates for the two points of intersection. To do this substitute each x value into either of the above equations. The points of intersection are located on both functions. Let us pick the equation that is easier to work with: y=9x-15

For x=7.5 For x=2

y=9x-15 y=9x-15

y=9(7.5)-15 y=9(2)-15

y=67.5-15 y=18-15

y=52.5 y=3

Therefore, the co-ordinates of the two points of intersection are (7.5,52.5) and (2,3).

When doing your homework, you may find a special type of problem which provides you with a quadratic function, and a tangent line, along with its slope. You will be requested to find the y-intercept. To do this write the equation of the line in slope y-intercept form. You'll need to work with two equations. (The equation of the quadratic function will already be given). Use a variable, such as b, for the y-intercept. Then equate the two equations and simplify. Next find a, b, and c. Once you've done this you'll be able to use the discriminant, b²-4ac, to find the y-intercept.

To make things easier let's go through an example:

A tangent line with a slope of 4 intersects the quadratic function, y=1/2x²+2x-8, at only one point. Find the y-intercept and write the equation of the tangent line in slope y-intercept form.

The first step is to write an equation for the line in slope y-intercept form, using a variable for the y-intercept. Let us use b to represent the y-intercept.

y=mx+b

y=4x+b The slope is 4.

Next equate the two equations, y=1/2x²+2x-8 & y=4x+b, and simplify:

1/2x²+2x-8=4x+b Equate.

1/2x²-2x-8-b=0 Simplify.

1/2x²-2x+(-8-b)=0 Place brackets around constant terms to isolate them. (Remember, b is a constant term)

Next, we must find the values of a, b, and c in the above simplified equation. This will allow us to work with the discriminant.

ax²+bx+c=0

1/2x²-2x+(-8-b)=0

Therefore, a=1/2, b=-2, and c=-8-b.

Now let us plug these values into the discriminant, b²-4ac. Remember the value of the discriminant for a tangent line is zero. Consequently, we can set up an equation that will allow us to isolate b, which represents the y-intercept of the tangent line.

a=1/2, b=-2, and c=-8-b.

b²-4ac=0

(-2)²-4(1/2)(-8-b)=0

4-2(-8-b)=0

4+16+2b=0

20+2b=0

2b=-20

b=-10

Therefore, the y-intercept of the tangent line is -10. The equation of the tangent line, in slope y-intercept is y=4x-10.

y=3x+5 and y=3x²-2x-4

First, equate the equations:

3x+5=3x²-2x-4

3x²-5x-9=0

Now, determine a, b, and c.

ax²+bx+c=0

3x²-5x-9=0

So, a=3, b=-5, and c=-9.

Now plug in the values of a, b, and c into the discriminant, b²-4ac.

b²-4ac

=(-5)²-4(3)(-9)

=25-12(-9)

=25+108

=133

Since the value of the discriminant is above zero, the line intersects the quadratic function at two points. Therefore, the line is a secant line. Now let us find the co-ordinates of the two points of intersection. To do this equate the two equations, and then simplify. Afterwards, find the values of x. Finally, plug in the values of x into either of the original equations to find y. Let's go through an example:

Find the co-ordinates of the two points of intersection for the following quadratic-linear system:

y=2x²-10x+15 and y=9x-15

2x²-10x+15=9x-15 First equate the two equations. (Make them equal).

2x²-19x+30=0 Simplify. Now you have a quadratic equation. Solve it.

2x²-4x-15x+30=0 In this case you can use factoring to solve the equation.

2x(x-2)-15(x-2)=0

(2x-15)(x-2)=0

x=7.5 or x=2

Now we must find the y-coordinates for the two points of intersection. To do this substitute each x value into either of the above equations. The points of intersection are located on both functions. Let us pick the equation that is easier to work with: y=9x-15

For x=7.5 For x=2

y=9x-15 y=9x-15

y=9(7.5)-15 y=9(2)-15

y=67.5-15 y=18-15

y=52.5 y=3

Therefore, the co-ordinates of the two points of intersection are (7.5,52.5) and (2,3).

When doing your homework, you may find a special type of problem which provides you with a quadratic function, and a tangent line, along with its slope. You will be requested to find the y-intercept. To do this write the equation of the line in slope y-intercept form. You'll need to work with two equations. (The equation of the quadratic function will already be given). Use a variable, such as b, for the y-intercept. Then equate the two equations and simplify. Next find a, b, and c. Once you've done this you'll be able to use the discriminant, b²-4ac, to find the y-intercept.

To make things easier let's go through an example:

A tangent line with a slope of 4 intersects the quadratic function, y=1/2x²+2x-8, at only one point. Find the y-intercept and write the equation of the tangent line in slope y-intercept form.

The first step is to write an equation for the line in slope y-intercept form, using a variable for the y-intercept. Let us use b to represent the y-intercept.

y=mx+b

y=4x+b The slope is 4.

Next equate the two equations, y=1/2x²+2x-8 & y=4x+b, and simplify:

1/2x²+2x-8=4x+b Equate.

1/2x²-2x-8-b=0 Simplify.

1/2x²-2x+(-8-b)=0 Place brackets around constant terms to isolate them. (Remember, b is a constant term)

Next, we must find the values of a, b, and c in the above simplified equation. This will allow us to work with the discriminant.

ax²+bx+c=0

1/2x²-2x+(-8-b)=0

Therefore, a=1/2, b=-2, and c=-8-b.

Now let us plug these values into the discriminant, b²-4ac. Remember the value of the discriminant for a tangent line is zero. Consequently, we can set up an equation that will allow us to isolate b, which represents the y-intercept of the tangent line.

a=1/2, b=-2, and c=-8-b.

b²-4ac=0

(-2)²-4(1/2)(-8-b)=0

4-2(-8-b)=0

4+16+2b=0

20+2b=0

2b=-20

b=-10

Therefore, the y-intercept of the tangent line is -10. The equation of the tangent line, in slope y-intercept is y=4x-10.

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