(FUNCTIONS) Need to find zeros & vertex

harem905

New member
Joined
Nov 12, 2013
Messages
6
Ok, so I'm 23 and taking a grade 11 functions and applications course and I'm seriously stuck on a question.

The question is f(x)=-2x^2-8x+5.

I worked as hard as I could and couldn't come to factors that added to 4 and multiplied to 2.5 so I went ahead and checked a graphing calculator. It gave me x=(-4.55) and x=(.55). My question there, is how I would ever come to those two answers on my own without the use of a calculator?

Now for the vertext, I'll show my work and the answer that I got (which I now know is wrong because of the graphing calculator).

-2x^2-8x+5
-2(x^2+4x)+5
-2(x^2+4x+4-4)+5
-2(x^2+4x+4)-8+5
-2(x+2)^2-3

Therefore, the point of vertex is (-2, -3).

The graphing calculator has the vertex at (-2, 13). I tried to get that number myself, but I'm not sure where I went wrong. Can anyone explain or show you work of how those numbers are attained? Thank you!
 

harem905

New member
Joined
Nov 12, 2013
Messages
6
I figured out how to get the vertex.

(-4.55+.55)/2=(-2)

Then I put the (-2) in the original equation.

-2(-2)^2-8(-2)+5
-2(4)+16+5
-8+16+5
13

Therefore the point of vertex is (-2, 13). Perfect. But now, how would I have/what do I need to know or do to get the zeros (x=(-4.55) and x=(.55)) on my own?
 

srmichael

Full Member
Joined
Oct 25, 2011
Messages
848
Ok, so I'm 23 and taking a grade 11 functions and applications course and I'm seriously stuck on a question.

The question is f(x)=-2x^2-8x+5.

I worked as hard as I could and couldn't come to factors that added to 4 and multiplied to 2.5 so I went ahead and checked a graphing calculator. It gave me x=(-4.55) and x=(.55). My question there, is how I would ever come to those two answers on my own without the use of a calculator?

Now for the vertext, I'll show my work and the answer that I got (which I now know is wrong because of the graphing calculator).

-2x^2-8x+5
-2(x^2+4x)+5
-2(x^2+4x+4-4)+5
-2(x^2+4x+4)-8+5
-2(x+2)^2-3

Therefore, the point of vertex is (-2, -3).

The graphing calculator has the vertex at (-2, 13). I tried to get that number myself, but I'm not sure where I went wrong. Can anyone explain or show you work of how those numbers are attained? Thank you!
Not sure where you are getting that you need two factors that add to 4 and multiply to 2.5. Fact of the matter is, this quadratic cannot be factored by normal factoring methods. If you need to solve for x (i.e. find the roots) then use the quadratic formula.

Now, regarding your attempt at completing the square to come up with the vertex formula...In your second to last step the -2 times the -4 = 8, not -8. You will then get -2(x + 2)² + 13.
 

Subhotosh Khan

Super Moderator
Staff member
Joined
Jun 18, 2007
Messages
18,183
Ok, so I'm 23 and taking a grade 11 functions and applications course and I'm seriously stuck on a question.

The question is f(x)=-2x^2-8x+5.

I worked as hard as I could and couldn't come to factors that added to 4 and multiplied to 2.5 so I went ahead and checked a graphing calculator. It gave me x=(-4.55) and x=(.55). My question there, is how I would ever come to those two answers on my own without the use of a calculator?

Now for the vertext, I'll show my work and the answer that I got (which I now know is wrong because of the graphing calculator).

-2x^2-8x+5
-2(x^2+4x)+5
-2(x^2+4x+4-4)+5
-2(x^2+4x+4)-8+5........... Incorrect ............that should be -2(x^2 + 4x + 4) + (-2)*(-4) + 5 = -2(x^2 + 4x + 4) + 8 + 5 = -2(x^2 + 4x + 4) + 13
-2(x+2)^2-3

Therefore, the point of vertex is (-2, -3).

The graphing calculator has the vertex at (-2, 13). I tried to get that number myself, but I'm not sure where I went wrong. Can anyone explain or show you work of how those numbers are attained? Thank you!
.
 

harem905

New member
Joined
Nov 12, 2013
Messages
6
Not sure where you are getting that you need two factors that add to 4 and multiply to 2.5. Fact of the matter is, this quadratic cannot be factored by normal factoring methods. If you need to solve for x (i.e. find the roots) then use the quadratic formula.

Now, regarding your attempt at completing the square to come up with the vertex formula...In your second to last step the -2 times the -4 = 8, not -8. You will then get -2(x + 2)² + 13.
Thank you for replying and pointing out that mistake, that makes things a little bit more clear. By suggesting I use the quadratic formula, do you mean x= (-b (+/-) the square root of b^2 -4ac)/2a? If so, since the b is already a negative, would I give it the opposite sign in the beginning of the formula or would it stay negative? Sorry if these are silly questions.
 

JeffM

Elite Member
Joined
Sep 14, 2012
Messages
3,258
Thank you for replying and pointing out that mistake, that makes things a little bit more clear. By suggesting I use the quadratic formula, do you mean x= (-b (+/-) the square root of b^2 -4ac)/2a? If so, since the b is already a negative, would I give it the opposite sign in the beginning of the formula or would it stay negative? Sorry if these are silly questions.
There are no silly questions.

First, using factoring to solve a quadratic is worth doing only if (a) you "see" a factoring without much effort, or (b) you have been told to solve using factoring. In the latter case, there is an implied promise that there is a relatively "easy" factoring. For most quadratic, there is not an "easy factoring. So I generally waste little time on factoring to solve quadratics.

Second, the methods that work for all quadratics are the quadratic formula and completing the square. The formula is usually easier so memorize it.

\(\displaystyle ax^2 + bx + c = 0 \implies x = \dfrac{- b \pm \sqrt{b^2 - 4ac}}{2a}.\)

Third, if b = - e, then - b = e.
 

Subhotosh Khan

Super Moderator
Staff member
Joined
Jun 18, 2007
Messages
18,183
There are no silly questions.

First, using factoring to solve a quadratic is worth doing only if (a) you "see" a factoring without much effort, or (b) you have been told to solve using factoring. In the latter case, there is an implied promise that there is a relatively "easy" factoring. For most quadratic, there is not an "easy factoring. So I generally waste little time on factoring to solve quadratics.

Second, the methods that work for all quadratics are the quadratic formula and completing the square. The formula is usually easier so memorize it.

\(\displaystyle ax^2 + bx + c = 0 \implies x = \dfrac{- b \pm \sqrt{b^2 - 4ac}}{2a}.\)

Third, if b = - e, then - b = e.
What happened to "d"?
 
Top