Without drawing the graph of the equation, determine how many points the parabola has in common with the x-axis and whether its vertex lies above, on, or below the x-axis.

y = –2x2 + x + 3

A.
no points in common; vertex below x-axis

B.

2 points in common; vertex above x-axis

C.

2 points in common; vertex below x-axis

D.

1 point in common; vertex on x-axis

Question

Without drawing the graph of the equation, determine how many points the parabola has in common with the x-axis and whether its vertex lies above, on, or below the x-axis.

 

    y = –2x2 + x + 3
	
	
 
A.
	no points in common; vertex below x-axis
	
 
B.
	

2 points in common; vertex above x-axis
	
 
C.
	

2 points in common; vertex below x-axis
	
 
D.
	

1 point in common; vertex on x-axis

whpalmer4 · Answer

First: does this parabola open up or down?

anonymous · Answer

will give medal

anonymous · Answer

@Bookworm14   @rose21  @undeadknight26

whpalmer4 · Answer

You can tell just by looking at the coefficient of the $x^2$ term.  If it is positive, then for sufficiently large values of $x$, $y$ will be an increasingly large positive number.  If it is negative, then again, as $x$ gets farther away from the y-axis, $y$ will again be an increasingly large positive number. 

In general, if you write the formula in the form $$y = ax^2+bx+c$$then if $a>0$, the parabola opens upward, like a bowl.  If $a< 0$ the parabola looks like an inverted bowl, and opens downward.

whpalmer4 · Answer

Next, you can use the discriminant of the quadratic equation to determine whether the parabola has 0, 1, or 2 points in common with the x-axis.

The discriminant, $\Delta$, is given by $\Delta = b^2-4ac$, which you may recognize as the quantity under the square root symbol in the formula for the solutions to a quadratic equation.

If $\Delta=0$, you have a perfect square, and will have just one solution.
If $\Delta > 0$, you have two real solutions.
If $\Delta < 0$, you have two complex solutions, in a conjugate pair $(a\pm bi)$

A real solution occurs where $y =0$.  Complex solutions don't have an obvious appearance on the graph, unlike real solutions.  A parabola which has only complex solutions appears suspended in midair above the x-axis (if opening upward) or hanging below the x-axis (if opening downward).

whpalmer4 · Answer

One more bit of parabola lore:

If you write the parabola in the standard quadratic form $y=ax^2+bx+c$ then you can find the x-coordinate of the vertex with the simple equation $$x = -\frac{b}{2a}$$and then find the $y$ value by plugging in that value of $x$.