Algebra 2 help!!! PLZ!!!!?

Question

For the function find the standard form, the vertex, line of symmetry, and the max or min value.......

f(x)= x^2+2x-5

Answer 1

Standard form is in the form f(x) = a(x - h)^2 + k

So complete the square:

f(x) = x^2 + 2x + ___ - 5
= x^2 + 2x + 1 - 5 - 1
= (x+1)^2 - 6

The vertex of a quadratic in standard form is (h, k), so (-1, -6).

The line of symmetry in this case is a vertical line through the vertex, x=-1

The max or min is the vertex, since the coefficient of x^2 is positive, the parabola opens upward and it's a min. The min value of f(x) is -6, when x=-1.

Answer 2

Hi I think I can help you.

1. Standardform

The functtion f(x) = x^2 + 2x - 5 is a parabula. To get the standard form you have to manipulate the equation in such a way that you eliminate the x-term. This is always possible by using the well knmown equation

( a + b )^2 = a^2 + 2*a*b + b^2

In our case we simply add on the right side of our equation +1 and subtract it at the same time, so in the end nothing has been changed.

f(x) = (x^2 + 2x + 1 ) -5 - 1 = (x + 1)^2 - 6

2. Vertex

You can read the coordinates of the vertex immidiatly from the standarform. The vertex has the coordinates

V(-1,-6)

You easily can proof this. Setting x = -1 brings f(x) to -6. This is the minimum value of f(x), because the square is always greater or equal to zero.

3. Line of symmetrie

This line goes through the vertex. The equation of the line of symmetrie is thertefore

x = -1 = constant for all values of f(x)

4. Max/Min-value

The vertex is with a parabula always the Max or the Min. Whether you have a maximum or a minimum you can either decide by drawing the function or bei using the first and second derivatives of f(x).

The first derivative is

f`(x) = 2*(x + 1) that means with f`(x) = 0 as the condition for an extrem value

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

The second derivative is

f``(x) = 2

If ``(x) is greater than zero than we have a minimum and
if f``(x) is below zero than we have a maximum.

Everything understood?

Answer 3

OK I see that the way to do these nowadays is to put them in standard form, which is slightly tricky but then gives you all of the other information you need very easily.

The standard form is: y = a(x - h)^2 + k

The vertex of this equation is (h, k) and the line of symmetry is therefore x=h.

I can see by looking at that this quadratic can be converted to standard form thusly:

x^2 + 2x - 5 = (x + 1)^2 - 6 (a = 1, h = -1, k = -6)

Therefore the vertex is (-1, -6), the line of symmetry is: x = -1

The coefficient of x^2 (a) is positive (in this case, 1) so the function has a minimum. The minimum is the value of 'y' at the vertex, which is therefore y = -6.

Hope this helps.

Maybe what you really need is help in completing the square. Ask, if so...

Answer 4

...

I havent done description work on graphsin awhile but the standard form im assuming is the shape of the graph. In this case its a quadratic parabola which is translated 5 units down the y axis and faces upwards.

The vertex will be at where f'(0)
f'(x) = 2x + 2
therefore at
f'(0) = 2
then sub 2 as the x-coordinate back into f(x) and you get the y coordinate

f(2) = 2^2 + 2(2) - 5 = 3
hence the vertex is at (2,3)

the line of symmetry is the same i think as the vertex x=2

the max or min value is something i havent done in awhile, cant really remember ahhaha i used to just use a graphics calc and plot it and look for min and max values

goodluck hope this helps

Answer 5

The standard form of function would be:
f(x)=x^2+2x-5
f(x)=x(x+2)-5
(-2,-5)
Line of symmetry would be at -1
Max at -6.
It's been a while since I've done this. Hope it helped or triggered some help.