Is the quadratic equation x^2 - 2x - 3 = 0
a parabola or two vertical lines when graphed?
since it is (x-3)(x+1) you know you have two roots. No straight line will cut the x-axis in more than one place (except y=0).
And since the coefficient of x^2 is nonzero, you know that it is in fact a parabola. I can't think of any quadratic that graphs to two lines.
The given quadratic equation is x^2 - 2x - 3 = 0.
To determine whether it graphs into a parabola or two vertical lines, we can analyze the discriminant (b^2 - 4ac) of the equation.
The quadratic equation is in the standard form ax^2 + bx + c = 0, where a = 1, b = -2, and c = -3.
The discriminant is calculated as follows:
Discriminant = b^2 - 4ac
= (-2)^2 - 4(1)(-3)
= 4 + 12
= 16
Since the discriminant (16) is a positive number, the quadratic equation x^2 - 2x - 3 = 0 graphs into a parabola.
To determine whether the quadratic equation x^2 - 2x - 3 = 0 represents a parabola or two vertical lines when graphed, we need to consider the discriminant of the equation.
The general form of a quadratic equation is ax^2 + bx + c = 0, where a, b, and c are constants. In this case, a = 1, b = -2, and c = -3.
The discriminant, denoted as Δ, is given by the formula Δ = b^2 - 4ac. For our equation, Ξ = (-2)^2 - 4(1)(-3).
Let's calculate the discriminant:
Δ = (-2)^2 - 4(1)(-3)
= 4 + 12
= 16
Now, let's analyze the discriminant's value:
1. If Δ > 0, the quadratic equation has two distinct real roots when graphed, and therefore the graph is a parabola that intersects the x-axis at two points.
2. If Δ = 0, the quadratic equation has one real root when graphed, and therefore the graph is a parabola that touches the x-axis at one point.
3. If Δ < 0, the quadratic equation has no real roots when graphed, and therefore the graph does not intersect the x-axis.
In our case, Δ = 16 which is greater than 0. Hence, the quadratic equation x^2 - 2x - 3 = 0 represents a parabola when graphed. It will intersect the x-axis at two distinct points.