Identifying the vertex and x-intercepts of a Quadratic Function?

Question

I don't understand this... why do i use (4/2)^2 in the next problem
Identifying the Vertex of a Quadratic Function
example
Describe the graph of f(x)= 2x^2+8x+7 and identify the vertex.
solution
write the quadratic function in standard form by completing the spare. Recall that the first step is to factor out any coefficient of x^2 that is not 1
f(x)=2x^2+8x+7
f(x)=2(x^2+4x)+7
f(x)=2(x^2 + 4x + 4 - 4) +7 ----------------------->because b=4, add and subtract (4/2)^2
f(x)=2(x^2 + 4x +4) - 2(4)+7
= 2(x+2)^2-1
From standard form, you can see that the graph of f is a parabola that opens upward with vertex (-2,-1)

Identify the vertex and x-intercepts(s).
#1
f(x)=x^2 - x + 5/4

I tried to factor it... (x- ) (x- ) but I could not figure out what times what is 5/4
the answer is the following but it want to know how to do this.
Vertex: (1/2, 20)
x-intercept: none
and a graph

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

Answer 1

An explanation of your example
f(x)= 2x² + 8x + 7

Think of the 7 as an after thought for now and set up the first steps of completing the square. Keep in mind that the 2 with the 2x² has to be moved to the outside... so that also has to apply to the 8x - it becomes a 4x inside the parenthesis.
f(x) = 2(x² + 4x + __ - __) + 7

What goes in the ___? Take the x term, the coefficient is b. Divide it by 2 and square it. In this case b is 4, 4 ÷ 2 is 2 and (2)² is 4. In this step, the __ is 4. [This is where you asked about the (4/2)². ]

f(x) = 2(x² + 4x + 4 - 4) + 7

Now, move the second -4 to the outside of the parenthesis, only now we have to consider that the outside coefficient isn't just a simple positive one. There's a 2 out there.
f(x) = 2(x² + 4x + 4) - 2(4) + 7

Think about factoring that parenthesis part into something that looks like a squared binomial: (x + __ )²
f(x) = 2(x + ___)² - 8 + 7
Hint: Use that b ÷ 2 term. DON'T SQUARE IT! In this case, b was 4, and 4 ÷ 2 is 2. Here the ___ is 2.

f(x) = 2(x + 2)² - 1
This is finally vertex form... y = a*(x - h)² + k where the vertex is at (h, k).

This parabola does open upward with a vertex of (-2, -1). The x-intercepts are at (-1.29, 0) and (-2.71, 0). The y-intercept is at 7.

I found the x-intercepts by solving the vertex form where y = 0.
y = 2(x + 2)² - 1
0 = 2(x + 2)² - 1
Add one to each side.

1 = 2(x + 2)²
Divide by 2 on each side.

1/2 = (x + 2)²
Take the square root on each side.

±√(1/2) = x + 2
Subtract 2 from each side.

-2 ±√(1/2) = x
x = -2 +√(1/2), -2 -√(1/2)
x = -1.29, -2.71

I think this is a lot less complicated than the quadratic formula... but you'll get the same thing! :-)
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For your problem #1... don't factor it... you have to complete the square...
f(x) = x² - x + 5/4

Think of the 5/4 as an after thought for now and set up the first steps of completing the square.
f(x) = (x² - x + __ - __) + 5/4

What goes in the ___? Take the x term, the coefficient is b. Divide it by 2 and square it. In this case b is -1, -1 ÷ 2 is -1/2 and (-1/2)² is 1/4. In this step, the __ is 1/4.

f(x) = (x² - x + 1/4 - 1/4) + 5/4
Now, move the second 1/4 to the outside of the parenthesis.

f(x) = (x² - x + 1/4) - 1/4 + 5/4
Think about factoring that parenthesis part into something that looks like a squared binomial: (x + __ )²

f(x) = (x + ___)² + 4/4
Hint: Use that b ÷ 2 term. DON'T SQUARE IT! In this case, b was -1, and -1 ÷ 2 is -1/2. Here the ___ is -1/2.

f(x) = (x² + (-1/2))² + 1
f(x) = (x² - 1/2 )² + 1
This is finally vertex form... y = a*(x - h)² + k where the vertex is at (h, k).

This parabola does open upward with a vertex of (1/2, 1), so there will be no x-intercepts. The y-intercept is at 5/4.

I got that there are no x-intercepts because the graph goes up from the y-value of 1. It can't cross the x-axis.

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For your problem #2...
f(x) = -x² + 2x + 5

Think of the 5 as an after thought for now and set up the first steps of completing the square. Keep in mind that the - with the -x² has to be moved to the outside... so the +2x becomes a -2x.
f(x) = -(x² - 2x +__ - __) + 5

What goes in the ___? Take the x term, the coefficient is b. Divide it by 2 and square it. In this case b is -2, -2 ÷ 2 is -1 and (-1)² is 1. In this step, the __ is 1.
f(x) = -(x² - 2x + 1 - 1) + 5

Now, move the second -1 to the outside of the parenthesis, only now we have to consider that the outside coefficient isn't just a simple positive one. There's a negative out there.
f(x) = -(x² - 2x + 1) - (-1) + 5

Think about factoring that parenthesis part into something that looks like a squared binomial: (x + __ )²
f(x) = -(x + ___)² + 1 + 5
Hint: Use that b ÷ 2 term. DON'T SQUARE IT! In this case, b was -2, and -2 ÷ 2 is -1. Here the ___ is -1.

f(x) = -(x + (-1))² + 6
f(x) = -(x - 1)² + 6
This is finally vertex form... y = a*(x - h)² + k where the vertex is at (h, k).

This parabola does open downward with a vertex of (1, 6). The x-intercepts are at (3.45, 0) and (-1.45, 0). The y-intercept is at 5.

I found these intercepts by solving the vertex form for y = 0:
y = -(x - 1)² + 6
0 = -(x - 1)² + 6
Subtract 6 from each side.

-6 = -(x - 1)²
Divide by the negative on each side.

6 = (x - 1)²
Take the square root of each side.

±√6 = x - 1
Add 1 to each side.

1 ± √6 = x
x = 1 + √6, 1 - √6
x = 3.45, -1.45

You'll get the same results with the quadratic formula... but I think this is easier.

I found the y-intercepts for both of these functions by looking at the c value from the y = ax² + bx + c form of the function.

There is a short-cut to completing the square... but I'm not so sure you should use it if you don't understand why and how completing the square works... maybe as a way to check if your answer is correct.

To get from y = ax² + bx + c to y = a*(x - h)² + k, you can use that the h = -b/2a and the k can be found by plugging in this h value for x into the y = ax² + bx + c form.

Thanks for the challenge... This took a while to type out. I hope it helped! :-)

Answer 2

The reason that you can't factor the problem is that it has no real solution. You have to solve it in the same way they did the example. The example is using the "complete the square method.

First, factor out any co-efficient for x^2. In this case, there isn't one, so you can skip this step.

Next, look at the coefficient for x: it is (-1). So you have to think, "What would the last term be if it were a perfect square x^2 -x + ??" The answer here is x^2 - x + 1/4 is a perfect square. So, you need to change your equation to look like that. (how to find the last term is always "divide the coefficient by 2, and then square it")

You can do that like this:

f(x) = x^2 - x +1/4 + 4/4

Bascially, it's the same equation, but you can now see the perfect square inside of it.

Now, just rewrite the perfect square to be:

f(x) = (x - 1/2)^2 + 4/4

of course, 4/4 = 1, so

f(x) = (x - 1/2)^2 + 1

This is what is called "vertex form." It's very easy to see the vertex. the x-coordinate is 1/2, because the there is (x-1/2) in the parentheses, and the y coordinate is 1, because there is 1 outside the parenthesis.

You need to look at some different "complete the square" problems and then try to go back at this. Or ask your teacher to explain how to complete the square again.

Try these pages:
http://www.purplemath.com/modules/sqrvertx.htm
http://mathforum.org/library/drmath/sets/select/dm_complete_square.html
http://www.biology.arizona.edu/BioMath/tutorials/Quadratic/CompletingtheSquare.html

Answer 3

f(x)=x^2 - x + 5/4
Complete the square by adding and subtracting the square of half the coefficient of x. Coefficient of x = -1, half of it = -1/2:
f(x) = [ x^2 - x + (1/2)^2 ] + [ 5/4 - (1/2)^2 ]
The first bracketed expression is (x - 1/2)^2. Hence:
f(x) = (x - 1/2)^2 + (5/4 - 1/4)
f(x) = (x - 1/2)^2 + 1 .......(1)
Hence the vertex is where x = 1/2, and f(x) or y = 1.
In other words, it's (1/2, 1) and not (1/2, 20).

There are no x-intercepts, as putting (1) = 0 gives:
(x - 1/2)^2 = -1.
As squared numbers are always positive, this quadratic does not have real roots. It cannot therefore be factorised.

Answer 4

Why you use (4/2)^2 :

Look at this general equation -
(x + a)^2 = x^2 + 2ax + a^2

Now look at the middle constant which is 2a.
You have to get from 2a to a^2.
So you take half of it (half of 2a is a).
Then you square it to get a^2.

So the rule is : take half of the constant which is
attached to x, and then square it.

With the equation f(x) = 2(x^2 + 4x) + 7, you take half of 4,
which is 2, and then square it, to get 4.

#1
Try again with your vertex equation, because you
should get (1/2, 1).

If you can't figure out how to factorise a quadratric,
or if it doesn't factorise easily, then you can either
use the quadratic formula, x = [-b ± sqrt(b^2 - 4ac)] / (2a)
or you can use the vertex form.

The vertex form for f(x) = x^2 - x + 5/4 is :
f(x) = (x - 1/2)^2 + 1
The x-intercepts occur at f(x) = 0, so,
(x - 1/2)^2 + 1 = 0
Subtract 1 from both sides :
(x - 1/2)^2 = -1
Take the square root of both sides :
x - 1/2 = ±sqrt(-1) = ±i
Add 1/2 to both sides :
x = 1/2 ± i
These are the imaginary x-intercepts.
There are no real x-intercepts.

#2
f(x) = -x^2 + 2x + 5
f(x) = -(x^2 - 2x) + 5
f(x) = -(x^2 - 2x + 1 - 1) + 5
f(x) = -(x^2 - 2x + 1) + 1 + 5
f(x) = -(x - 1)^2 + 6
Therefore, vertex is at (1, 6)

For x-intercepts, let -(x - 1)^2 + 6 = 0
Therefore, (x - 1)^2 = 6
x - 1 = ±sqrt(6)
x = 1 ± sqrt(6)

Answer 5

f(x) = x^2 - x + 5/4
f(x) = x^2 - x + (1/2)^2 + 5/4 - (1/2)^2 . . add and subtract (1/2)^2
f(x) = y = (x - 1/2)^2 + 5/4 -1/4 . . let f(x) = y
y = (x - 1/2)^2 + 1
y - 1 = (x - 1/2)^2 . . . this is the equation of parabola
. . . . vertex (1/2, 1) parabola is facing up
. . . when x = 0 , y = 5/4

f(x) = -x^2 + 2x +5 . . . . let f(x) = y
y = -x^2 + 2x +5
y = -( x^2 - 2x + 1) +6 . . . . inside the bracket is -1 + 6 = 5
y = - (x - 1)^2 + 6
y - 6 = - (x - 1)^2 . . . . . . this is the equation of parabola
. . . . vertex (1, 6) . . . . parabola is facing down
. . . . when x = 0 . . . y = 5

Answer 6

#1
f(x) = x^2 - x + 5/4
y = x^2 - x + 1/4 - 1/4 + 5/4
y = (x - 1/2)^2 + 4/4
y = (x - 1/2)^2 + 1
vertex (1/2, 1), opens up so no x intercepts
x = 1/2 + i, 1/2 - i

# 2
f(x) = - x^2 + 2x + 5
y = - (x^2 - 2x) + 5
y = - (x^2 - 2x + 1 - 1) + 5
y = - (x - 1)^2 + 6
vertex at (1,6), opening down
y = - [(x - 1)^2 - 6]
y = - (x - 1 - √6)(x - 1 + √6)
x = (1 + √6), (1 - √6)

Answer 7

set x equal to 0 b is (0,1) a is (0, 13)