How would I find the vertex of the parabola y = x^2 – 2x + 4 ??

Question

can some one show me the method please

Answer 1

if parabola is in form ax^2 +bx+c then the x cordinate of the vertex is equal to: -b/2a.

In this example vertex = -(-2)/2(1) = 1 = x
y = 1^2 -2(1) + 4 = 3

The coordinate of the vertex is then (1, 3).

Answer 2

actual this question is approximately factorising: y = x^2 - 2x - 8 = (x + 2)(x - 4) so which you resolve for the graph, with the graph reducing the x-axis on the standards -2 and four. now the vertex of the graph is the midpoint of the graph, which runs contained in direction of the midpoint of the two factors 2 and -4. by means of calculation, the midpoint of two and -4 could be got here across by means of finding the area: 4 - (-2) = 6 and then dividing by means of 2 6 / 2 = 3 and then including this 3 to the left factor that's -2 or subtracting it from the nicely suited factor that's 4 the two way you get a cost of a million. for this reason the vertex passes contained in direction of the x-axis on the factor a million. to locate the y fee, you merely substitute a million into the unique equation: y = x^2 - 2x - 8 supplying you with: (a million)^2 - 2(a million) -8 = a million - 2 -8 = -9 for this reason the respond is (c) (a million, -9)

Answer 3

For an equation of the form y = (x - h)^2 + k, the vertex is at (h, k) ........(1)
To complete the square and achieve this form from your equation, you add to each side of the equation the square of half the coefficient of x.
y = x^2 - 2x + 4
The square of half -2 is (-1)^2 = 1.
y + 1 = (x^2 - 2x + 1) + 4
Subtract 1 from each side, and factorise the square in brackets:
y = (x - 1)^2 + 3
Comparing this equation with (1) above, you see that the vertex is (1, 3).

Answer 4

I remember that the Max or Min point of a parabola is in the form:
( -b/(2a), -b²/(4a) + c ) =

( -(-2)/(2*1), -(-2)²/(4*1) + 4 ) =
(2/2, -4/4 + 4) =
(1, -1 + 4) =
(1,3)

Answer 5

The vertex lies on the line x = -b/2a = 2/2=1
So y = 1-2+4= 3
So vertex is at (1,3)

Answer 6

The question you are asking is when does the increase in y go to zero, so:

dy/dx=2x-2

So 0=2x-2, when x=1, y=0. At the point then 0, 4, would be your vertex by plugging zero into the original equation.

Answer 7

mradigan747 has the right idea, but his calculation is incorrect.
1^2 - 2*1 +4 = 3.
(1,3) is your vertex

Answer 8

y = (x² - 2x + 1) - 1 + 4
y = (x - 1)² + 3
vertex is when x = 1, y = 3
OR
f(x) = x² - 2x + 4
f `(x) = 2x - 2 = 0 for turning point
x = 1 for turning point.
f(1) = y = 3
vertex at (1,3) as above.

Answer 9

Complete the square.

y = x² - 2x + 4
y = (x² - 2x + 1) + 3
y - 3 = (x - 1)²

The vertex is (1, 3).

Answer 10

x= -b/2a(this is the formula you use to find the axis of symmetry)

x= -(-2)/2(1)
x=2/2
x=1

substitute it in
y=x^2-2x+4
y=(1)^2-2(1)+4
y=1-2+4
y= -1+4
y=3

vertex=(1,3)