Please check for me.Without drawing the graph of the given equation determine
(a)how many x-intercepts the parabola has
(b)whether it vertex lies above, below or on the axis.
1. y=x^2-5x+6
I use the determinant
sqrt b^2 -4ac
(-5)^2-4(1)(6))=1
There are two real numbers
The parabola has two X- intercepets
The parabola opens downward and its vertex lies above the axis.
2.Y=-X^2+2X-1
(2)^2-4(-1)(-1)=0
There is one real number
one x-intercept
The parabola opens upward and its vertex lies on the axis.
Thanks Much!!!
119
To find the number of x-intercepts of a parabola without drawing the graph, you can use the discriminant of the quadratic equation.
(a) The discriminant, obtained by calculating the square root of b^2 - 4ac, tells us how many real solutions there are for the quadratic equation. If the discriminant is positive, there are two real solutions (x-intercepts), if it is zero, there is one real solution, and if it is negative, there are no real solutions.
(b) To determine whether the vertex lies above, below, or on the x-axis, you can consider the coefficient of x^2. If it is positive, the parabola opens upward and the vertex lies below the x-axis. If it is negative, the parabola opens downward and the vertex lies above the x-axis. If the coefficient is zero, the equation is not a quadratic and doesn't represent a parabola.
Let's apply this to the given equations:
1. y = x^2 - 5x + 6
a = 1, b = -5, c = 6
Discriminant: √((-5)^2 - 4(1)(6)) = √(25 - 24) = √1 = 1
Since the discriminant is positive, there are two real solutions (x-intercepts).
The coefficient of x^2 is positive, so the parabola opens upward. Therefore, the vertex lies below the x-axis.
2. y = -x^2 + 2x - 1
a = -1, b = 2, c = -1
Discriminant: √(2^2 - 4(-1)(-1)) = √(4 - 4) = √0 = 0
Since the discriminant is zero, there is one real solution (x-intercept).
The coefficient of x^2 is negative, so the parabola opens downward. Therefore, the vertex lies above the x-axis.
I hope this clarifies the process! Let me know if you have any more questions.