Question

y=2x^2-8x+7
i have to find the axis of symmetry
the coordinates of the vertex
and tell whether the vertex is a maximum or a minimum

Answer 1

-b/2a ia axis of symmetry

--8/2.2

8/4 = 2

Answer 2

ia is stupidity for "is" ...

Answer 3

haha thank you

Answer 4

\[y = 2x^2 - 8x + 7\]
\[ = 2(x^2 - 4x) + 7\]
\[= 2(x^2 -4x + 4) -8 + 7\]
\[\implies y +1 = 2(x-2)^2\]

Answer 5

So we have a parabola which is shifted 1 unit up in y and 2 units forward in x and grows twice as fast as the standard parabola.

Answer 6

Err shifted 1 unit down in y, sorry.

Answer 7

oh ok u got me confused 4 a sec

Answer 8

what part of the problem are u helping with

Answer 9

Finding the vertex, and the axis of symmetry.

Answer 10

Actually this form of the function answers all of those questions.

Answer 11

i did not learn it that way can u help me another way by chance

Answer 12

Since the coefficient of x is positive, the graph is opening upward. Therefore the vertex will be a min. The location of the vertex can be found from the shift in x and y of the function from (0,0). In this case we shifted down 1 unit and 2 units to the right.

\[y-y_0 = m(x-x_0)^2\]
Is the standard form for a parabola. Where \(x_0,y_0\) are the x and y points of your vertex. And m is the scaling and directional coefficient.

So we have \(y - (-1) = 2(x - 2)^2\) which tells us the vertex is at (2,-1)

And since the graph is opening upward the line x = 2 will be the line of symmetry (the line going up/down that goes through the vertex)

Answer 13

thank you soooooo much