Question 147230
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 {{{D=b^2-4ac}}}. 


From {{{x^2-5x+6}}} we can see that {{{a=1}}}, {{{b=-5}}}, and {{{c=6}}}



{{{D=b^2-4ac}}} Start with the discriminant formula



{{{D=(-5)^2-4(1)(6)}}} Plug in {{{a=1}}}, {{{b=-5}}}, and {{{c=6}}}



{{{D=25-4(1)(6)}}} Square {{{-5}}} to get {{{25}}}



{{{D=25-24}}} Multiply {{{4(1)(6)}}} to get {{{(4)(6)=24}}}



{{{D=1}}} Subtract {{{24}}} from {{{25}}} to get {{{1}}}



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.


{{{x=-b/(2a)}}}


From the equation {{{y=x^2-5x+6}}} we can see that a=1 and b=-5


{{{x=(--5)/(2*1)}}} Plug in b=-5 and a=1



{{{x=5/(2*1)}}} Negate -5 to get 5



{{{x=(5)/2}}} Multiply 2 and 1 to get 2



So the x-coordinate of the vertex is {{{x=5/2}}}. Lets plug this into the equation to find the y-coordinate of the vertex.



{{{y=x^2-5x+6}}} Start with the given polynomial



{{{y=(5/2)^2-5(5/2)+6}}} Plug in {{{x=5/2}}}



{{{y=25/4-5(5/2)+6}}} Square {{{5/2}}} to get {{{25/4}}}



{{{y=25/4-25/2+6}}} Multiply 5 by {{{5/2}}} to get {{{25/2}}}



{{{y=-1/4}}} Now combine like terms



So the vertex is *[Tex \LARGE \left(\frac{5}{2},-\frac{1}{4}\right)]



Since the y-coordinate of the vertex is {{{y=-1/4}}}, this means that the vertex is below the x-axis