Remember that a quadratic equation is a parabola. The equation is of the type y = Ax^2 + Bx + C
A linear equation is a straight line. The equation is of the type y = MX + N
The soluction of that system is Ax^2 + Bx + C = MX + N
=> Ax^2 + (B-M)x + (C-N) = 0
That is a quadratic equation.
A quadratic equation may have 0, 1 or 2 real solutions. Those are all the possibilitis.
So you must select 0, 1 and 2.
You can also get to that conclusion if you draw a parabola and figure out now many point of it you can intersect with a straight line.
You will realize that depending of the straight line position it can intersect the parabola in none point, or one point or two points.
What’s the yintercept of the function?
Answer:
11
Step-by-step explanation:
Let x be the unknown number
2x+5 = 27
Subtract 5 from each side
2x+5-5 = 27-5
2x = 22
Divide each side by 2
2x/2 = 22/2
x = 11
The number is 11
First, you subtract the equations. (3x^2-2x+5) - (x^2+3x-2) so you get 3x^2 - 2x + 5 - x^2 - 3x + 2. This leads you to 2x^2 - 5x + 7. If you multiply that by 1/2 x^2, you get x^4 - 5/2(x^3) + 7/2(x^2)
