find the intersection between
parabola: y=x^2 + 3x - 4
and
line: y=5x + 11
Thankyou :)
Two intersections or none are possible
5x + 11 = x^2 + 3x - 4
x^2 - 2x -15 = 0
(x-5)(x+3) = 0
x = +5 then y = +36
x = -3 then y = - 4
just one intersection is possible as well
Find the intersection between parabola y=2x^2+3x+2 and line y=2x+3
To find the intersection points between a parabola and a line, you need to solve the system of equations formed by setting the equations of the parabola and line equal to each other.
In this case, we have the equations:
1. Parabola: y = x^2 + 3x - 4
2. Line: y = 5x + 11
To find the intersection points, we substitute the equation of the line into the equation of the parabola:
x^2 + 3x - 4 = 5x + 11
Now, we have a quadratic equation. We can simplify it by moving all terms to one side of the equation:
x^2 + 3x - 5x - 4 - 11 = 0
x^2 - 2x - 15 = 0
To solve this quadratic equation, we can factor it or use the quadratic formula. Let's use the quadratic formula to find the values of x:
The quadratic formula is given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
For the equation x^2 - 2x - 15 = 0, our coefficients are:
a = 1, b = -2, and c = -15
Substituting these values into the quadratic formula:
x = (-(-2) ± √((-2)^2 - 4(1)(-15))) / (2(1))
x = (2 ± √(4 + 60)) / 2
x = (2 ± √(64)) / 2
x = (2 ± 8) / 2
Now, we have two possible solutions for x:
1. x = (2 + 8) / 2 = 5
2. x = (2 - 8) / 2 = -3
Now, substitute these x-values back into either the equation of the parabola or the line to find the corresponding y-values.
For x = 5:
Parabola: y = (5)^2 + 3(5) - 4 = 25 + 15 - 4 = 36
Line: y = 5(5) + 11 = 25 + 11 = 36
For x = -3:
Parabola: y = (-3)^2 + 3(-3) - 4 = 9 - 9 - 4 = -4
Line: y = 5(-3) + 11 = -15 + 11 = -4
Therefore, the intersection points between the parabola and the line are (5, 36) and (-3, -4).