help me please?

Question

I have this problem that I can't seem to figure out. Can you help me?

Without drawing the graph of the given equation, determine
(a) how many x-intercepts the parabola has
(b) whether its vertex lies above below or on the x-axis.

y = -x^2 + 2x - 1

Answer 1

To find the x - intercepts, set the equation equal to zero and solve.
-x^2 + 2x - 1 = 0
Factor out and divide by the negative to make life easier for you, leaving...
x^2 - 2x + 1 = 0
(x - 1)(x - 1) = 0
x - 1 = 0 and x - 1 = 0
x = 1 and x = 1
Thus the only x - intercept is (1, 0).
Now, since there is only one x - intercept, it _must_ be the vertex, therefore, the vertex lies ON the x - axis.
Hope this helps.

Answer 2

Hello

Because there is a negative in front of the x^2 it opens downward. The y - intercept is -1 and the vertex is found by using -b/(2*a) or -2/(2*-1) = 1 putting this x value into the equation gives us the y value of the vertex or

y = -(1)^2 + 2*1 -1 = 0

So the vertex is at (1,0) this is on the x-axis is max point and is on the x-axis and since it opens downward there is only one x-intercept.

Answer 3

y = -x^2 + 2x - 1

to find the intersects in x-axis

y = 0

when y = 0

-x^2 + 2x - 1 = 0
now factorize -1

-1(x^2 - 2x + 1) = 0
x^2 -2x + 1 = 0

(x - 1)(x - 1) = 0
x = 1

so the min or maximum poin touches x = 1;

b) now to find whether the point obtained above is maximum point or minimum point.

dy/dx = -2x + 2

d2y/dx2 = -2 (which is < 0)

this means the point is a max point.
Therefore all the points are below x-axis.

if you still don't understand you should just sketch a graph to help.

Answer 4

the parabola would cross the x-axes when y=0
therefore let the equation equal to zero

0=-x^2+2x-1

0=(-x+1)(x-1)
-x+1=0, x-1=0
x=1, x=1
both equations is the same answer which means x=1, it only crosses the x-axes one time.

Answer 5

y = -x^2 + 2x - 1
First multiplying through by - 1 foe clarity
-y = x^2 - 2x + 1
-y = (x - 1)^2
y = - (x - 1)^2
a) There is one x-intercept, the vertex
b) The vertex lies at (1,0)