SOLUTION: I need help, if you may. It says without drawing the graph of the given equation, determine: How many x-intercepts the parabola had and whether its vertex lies above or below or on

Question 147230This question is from textbook algebra
: I need help, if you may. It says without drawing the graph of the given equation, determine: How many x-intercepts the parabola had and whether its vertex lies above or below or on the x-axis.
y=x^2-5x+6
Thank you=) This question is from textbook algebra

Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
To figure out how many x-intercepts there are, we need to find out how many solutions there are. So let's use the discriminant formula .

From we can see that , , and

Start with the discriminant formula

Plug in , , and

Square to get

Multiply to get

Subtract from to get

Since the discriminant is greater than zero, this means that there are two real solutions. So there are two distinct x-intercepts.

As a quick note, since there are two distinct x-intercepts, this means that the vertex cannot be on the x-axis. So it is either above or below the x-axis.
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To find out if the vertex is above or below the x-axis, we need to find the y-coordinate of the vertex. To do that, we need to find the x-coordinate of the vertex.

So let's use this formula to find the x-coordinate of the vertex.

From the equation we can see that a=1 and b=-5

Plug in b=-5 and a=1

Negate -5 to get 5

Multiply 2 and 1 to get 2

So the x-coordinate of the vertex is . Lets plug this into the equation to find the y-coordinate of the vertex.

Start with the given polynomial

Plug in

Square to get

Multiply 5 by to get

Now combine like terms

So the vertex is

Since the y-coordinate of the vertex is , this means that the vertex is below the x-axis

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