the intersecting line and parabola
Example 1
The line y = 2x - 1 intersects a parabola y = (x-3)^2+2. Find the intersecting points. Graphs do not need to be drawn. The equations must be arranged algebraically to the point that a quadratic equation is formed but will be solved with the graphics calculator or DESMOS. There is a formula (the quadratic formula) useful for M/E examples but for now this is sufficient.
rearranging the expressions to create a quadratic equation
1) make the two equations equal each other.
2) move all x terms over to the left hand side, changing their signs as you go
3) move all constant terms (numbers) to the left hand side, changing their signs as you go.
Positive becomes negative and negative becomes positive.
Once the quadratic equation has been achieved, enter the coefficients of x , x^2 and the constant into the array in the calculator.
The calculator will provide both values of x but you will have to work out what y is by substituting x into an original equation, it doesn't matter which one but one will be usually be easier than the other, probably the straight line equation.
Now let's go through the calculations step bt step.
1) Now enter those values into the calculator: In equation mode select polynomial, 2 degrees and enter into the array, 1, -8 and 12 select solve and write down the x values.
2) Substitute the x values into the line equation to find the y values.
3) Give your answer in coordinate form.
So one location is at x = 2 and the other location is at x = 6
Therefore because y = 2x - 1
when x = 2, y = 2 x 2 - 1
y = 3
and when x = 6, y = 2 x 6 - 1
y = 11
So point A is at (2,3) and point B is at (6,11)
Line and parabola example 2
The line y = 2x + 3 intersects a parabola y = 1/2(x-2)^2+1. Find the intersecting points. Graphs do not need to be drawn. The equations must be arranged algebraically to the point that a quadratic equation is formed but will be solved with the graphics calculator or DESMOS.
Rearranging The Expressions To Create A Quadratic Equation
1) make the two equations equal each other.
2) move all x terms over to the left hand side, changing their signs as you go
3) move all constant terms (numbers) to the left hand side, changing their signs as you go.
Positive becomes negative and negative becomes positive.
Once the quadratic equation has been achieved, enter the coefficients of x , x^2 and the constant into the array in the calculator.
The calculator will provide both values of x but you will have to work out what y is by substituting x into an original equation, it doesn't matter which one but one will be usually be easier than the other, probably the straight line equation.
Now let's go through the calculations step by step.
2) move all x terms over to the left hand side, changing their signs as you go
3) move all constant terms (numbers) to the left hand side, changing their signs as you go.
Positive becomes negative and negative becomes positive.
Once the quadratic equation has been achieved, enter the coefficients of x , x^2 and the constant into the array in the calculator.
The calculator will provide both values of x but you will have to work out what y is by substituting x into an original equation, it doesn't matter which one but one will be usually be easier than the other, probably the straight line equation.
Now let's go through the calculations step by step.
1) Now enter those values into the calculator: In equation mode select polynomial, 2 degrees and enter into the array, 1, -8 and 0, select solve and write down the x values.
2) Substitute the x values into the line equation to find the y values.
3) Give your answer in coordinate form.
2) Substitute the x values into the line equation to find the y values.
3) Give your answer in coordinate form.
So one location is at x = 8 and the other location is at x = 0
Therefore because y = 2x + 3
when x = 8, y = 2 x 8 + 3
y = 19
and when x = 0, y = 2 x 0 + 3
y = 3
Therefore, point A is at (0,3) and point B is at point (8,19)
Therefore because y = 2x + 3
when x = 8, y = 2 x 8 + 3
y = 19
and when x = 0, y = 2 x 0 + 3
y = 3
Therefore, point A is at (0,3) and point B is at point (8,19)
Line and parabola example 3
The line y = 2x + 6 intersects a parabola y = 1/4(x-4)^2+2. Find the intersecting points. Graphs do not need to be drawn. The equations must be arranged algebraically to the point that a quadratic equation is formed but will be solved with the graphics calculator or DESMOS.
Rearranging the expressions to form a quadratic equation
1) make the two equations equal each other.
2) move all x terms over to the left hand side, changing their signs as you go
3) move all constant terms (numbers) to the left hand side, changing their signs as you go.
Positive becomes negative and negative becomes positive.
Once the quadratic equation has been achieved, enter the coefficients of x , x^2 and the constant into the array in the calculator.
The calculator will provide both values of x but you will have to work out what y is by substituting x into an original equation, it doesn't matter which one but one will be usually be easier than the other, probably the straight line equation.
Now let's go through the calculations step by step.
1) make the two equations equal each other.
2) move all x terms over to the left hand side, changing their signs as you go
3) move all constant terms (numbers) to the left hand side, changing their signs as you go.
Positive becomes negative and negative becomes positive.
Once the quadratic equation has been achieved, enter the coefficients of x , x^2 and the constant into the array in the calculator.
The calculator will provide both values of x but you will have to work out what y is by substituting x into an original equation, it doesn't matter which one but one will be usually be easier than the other, probably the straight line equation.
Now let's go through the calculations step by step.
1) Now enter those values into the calculator: In equation mode select polynomial, 2 degrees and enter into the array, 1, -16 and 0, select solve and write down the x values.
2) Substitute the x values into the line equation to find the y values.
3) Give your answer in coordinate form.
2) Substitute the x values into the line equation to find the y values.
3) Give your answer in coordinate form.
So one location is at x = 16 and the other location is at x = 0
Therefore because y = 2x + 6
when x = 16, y = 2 x 16 + 6
y = 38
and when x = 0, y = 2 x 0 + 6
y = 6
Therefore, point A is at (0,6) and point B is at (16,38)
Therefore because y = 2x + 6
when x = 16, y = 2 x 16 + 6
y = 38
and when x = 0, y = 2 x 0 + 6
y = 6
Therefore, point A is at (0,6) and point B is at (16,38)
Line And Parabola Example 4
with decimal solution
The line y = 1/4x + 6 intersects a parabola y = 1/2(x-4)^2+2. Find the intersecting points. Graphs do not need to be drawn. The equations must be arranged algebraically to the point that a quadratic equation is formed but will be solved with the graphics calculator or DESMOS.
Rearranging the expressions to form a quadratic equation
1) make the two equations equal each other.
2) move all x terms over to the left hand side, changing their signs as you go
3) move all constant terms (numbers) to the left hand side, changing their signs as you go.
Positive becomes negative and negative becomes positive.
Once the quadratic equation has been achieved, enter the coefficients of x , x^2 and the constant into the array in the calculator.
The calculator will provide both values of x but you will have to work out what y is by substituting x into an original equation, it doesn't matter which one but one will be usually be easier than the other, probably the straight line equation.
Now let's go through the calculations step by step.
1) make the two equations equal each other.
2) move all x terms over to the left hand side, changing their signs as you go
3) move all constant terms (numbers) to the left hand side, changing their signs as you go.
Positive becomes negative and negative becomes positive.
Once the quadratic equation has been achieved, enter the coefficients of x , x^2 and the constant into the array in the calculator.
The calculator will provide both values of x but you will have to work out what y is by substituting x into an original equation, it doesn't matter which one but one will be usually be easier than the other, probably the straight line equation.
Now let's go through the calculations step by step.
1) Now enter those values into the calculator: In equation mode select polynomial, 2 degrees and enter into the array, 1, -7.5 and 8, select solve and write down the x values to 2 decimal places.
2) Substitute the x values into the line equation to find the y values.
3) Give your answer in coordinate form.
2) Substitute the x values into the line equation to find the y values.
3) Give your answer in coordinate form.
So one location is at x = 6.21 and the other location is at x = 1.29
Therefore because y = -1/4x + 6
when x = 6.21, y = -1/4 x 6.21 + 6
y = 4.45
and when x = 1.29, y = -1/4 x 1.29 + 6
y = 5.68
Therefore, point A is at (1.29,5.68) and point B is at (6.21,4.45)
Therefore because y = -1/4x + 6
when x = 6.21, y = -1/4 x 6.21 + 6
y = 4.45
and when x = 1.29, y = -1/4 x 1.29 + 6
y = 5.68
Therefore, point A is at (1.29,5.68) and point B is at (6.21,4.45)
SYSTEMS OF EQUATIONS BOOKLET
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mahobe_l2_systems_of_equations.pdf | |
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