Question 147536
How many x-intercepts does the parabola have? 


To find this out, we need to find out how many solutions there are. So we need to use the discriminant formula {{{D=b^2-4ac}}}.



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



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



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



{{{D=4-4(-1)(-1)}}} Square {{{2}}} to get {{{4}}}



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



{{{D=0}}} Subtract {{{4}}} from {{{4}}} to get {{{0}}}



Since the discriminant is equal to zero, this means that there is only one real solution.


Since there is only one real solution, there is only one x-intercept. Because there is only one x-intercept, this means that the vertex must lie on the x-axis. However, this may not be so obvious. So let's find out where the vertex is.




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Where does the vertex lie?


To find out if the vertex is above or below the x-axis, we need to find the y-coordinate of the vertex. However, we first need to find the x-coordinate of the vertex.


To find the x-coordinate of the vertex, use this formula: {{{x=(-b)/(2a)}}}.



{{{x=(-b)/(2a)}}} Start with the given formula.



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



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



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



{{{x=1}}} Divide.



So the x-coordinate of the vertex is {{{x=1}}}. Note: this means that the axis of symmetry is also {{{x=1}}}.



Now that we know the x-coordinate of the vertex, we can use it to find the y-coordinate of the vertex.



{{{y=-x^2+2x-1}}} Start with the given equation.



{{{y=-(1)^2+2(1)-1}}} Plug in {{{x=1}}}.



{{{y=-1+2(1)-1}}} Square 1 to get 1.



{{{y=-1+2-1}}} Multiply 2 and 1 to get 2.



{{{y=0}}} Combine like terms.



So the y-coordinate of the vertex is {{{y=0}}}, which means that the the vertex is on the x-axis. So this confirms our original claim.