I dont get it can somebody please help me7?

Question

Find the vertex either by writing the parabola in standard form or using the vertex formula.

f(x) = x^2-16x +3

Answer 1

f(x)
= x^2-16x +3
= x^2-16x+64 - 64+3
= (x-8)^2 - 61
Vertex: (8, -61)

Answer 2

It is easy to forget things like the vertex formula. Learn how to complete the square by squaring half of -16 (coefficient of x term) getting 64 and then add and subtract 64 to the right side of the equation. Be sure to leave the 3 as is. Note adding and subtracting the same number does not change the value of the function, just is form.

f(x) = x^2 = 16x +64 -64 +3.

Now factor the first three terms which form a perfect square so the factoring is easy, and combine the last two terms to get

f(x) = (x-8)^2 - 61

This is called Standard Form. The reason for this is that when written as a perfect square +/= a constant you can read off the coordinates of the vertex.

Since (x-8)^2 is either 0 or positive the parabola open up and y=f(x) rises higher and higher as x gets larger.

So the vertex of this parabola is its lowest point. And that will occur when (x-8)^2 takes on its smallest value which is 0 (remember it is positive the rest of the time from being squared).

But look at (x-8)^2. It will be zero when you set x=8 leaving
-61 for f(8).

So the vertex is at (8,-61).

Now if you like to remember formulas instead of proccesses, the vertex formula says that the x-coordinate of the vertex is at

Xv = -b/2a where a is the coefficient of x^2, and b of x.

So let's check our work, Xv = -(-16)/2 = 8 which is what we got above.

How can we find the y coordinate of the vertex? Calculate f(8). Notice how easy this will be in the standard form. Again we get -61.

Answer 3

Find the vertex of
f(x) = x^2-16x +3

write this in the vertex form
which is
f(x) = a(x-m)^2+n

expanding this to get
a(x^2+m^2 -2xm)+n = ax^2 +am^2 -2amx +n=
ax^2 -2amx +(am^2+n) = x^2-16x+3
by identification we find
a=1 m=8 and n = 3-am^2 = 3-8^2 =-61
vertex form is
(x-8)^2 -61
the vertex is at point (8, -61)

Answer 4

Xv= 8 Yv= 64-128+3=-61
y=(x-8)^2 -61

Answer 5

y = x^2 -16x +3
y-3 +(16/2)^2 = x^2 -16x +(16/2)^2
y-3 + 64 = (x-8)^2
y + 61 = (x-8)^2

vertex = (8,-61)