Why Does Completing the Square Yield Complex Numbers in This Equation?

[NoLimits]

Hi,

I am having trouble with completing the square in this question. I've looked at several online videos and I am still confused as to why I am not getting an answer that makes sense. I keep ending up with complex numbers and I am not sure what the next step to take is.

Homework Statement

Determine the equation of the tangent to the curve defined by f(x)=x^2-6x+14 at (1, 9) and then sketch it.

Homework Equations

f(x)=x^2-6x+14

The Attempt at a Solution

f'(1) = lim_h→0 \frac{f(x+h)-f(x)}{h}
= lim_h→0 \frac{[(1+h)^2-6(1+h)+14]-[1^2-6(1)+14]}{h}
= lim_h→0 \frac{[1+h+h^2+h+14]-[1-6+14]}{h}
= lim_h→0 \frac{[1+2h+h^2+14]-[1-6+14]}{h}
= lim_h→0 \frac{2h+h^2-6}{h}
= lim_h→0 (2+h-6)
= 2+(0)-6
= -4

Determine equation of tangent line:

y = mx+b
9 = (-4)(-1) + b
9 = -4 + b
4 + 9 = b
b = 13

Therefore, the equation of the tangent to the curve is y= -4x + 13.

Completing the square:

x^2 - 6x + 14 = 0
x^2 - 6x + 9 + 14 - 9 = 0
(x-3)^2 + 14 - 9 = 0
(x-3)^2 + 5 = 0
(x-3)^2 = -5

<br /> \sqrt{(x-3)^2} = +- \sqrt{-5}<br />

x-3 = \sqrt{5}i
x = \sqrt{5}i + 3

and

x-3 = -\sqrt{5}i
x = -\sqrt{5}i + 3

 

[Saitama]

Hi NoLimits!

= lim_h→0 \frac{[(1+h)^2-6(1+h)+14]-[1^2-6(1)+14]}{h}
= lim_h→0 \frac{[1+h+h^2+h+14]-[1-6+14]}{h}

This doesn't look right. Can you show how do you get $(1+h)^2-6(1+h)+14$ equal to $1+h+h^2+h+14$?

 

[NoLimits]

Oops. I messed up there, however the end result of the equation (-4) is still the same. The line should've read:

1 + 2h + h^2 - 6 - 6h + 14

 

[Mentallic]

By sheer luck you arrived at the correct answer (even after making many mistakes).

= lim_h→0 \frac{[(1+h)^2-6(1+h)+14]-[1^2-6(1)+14]}{h}
= lim_h→0 \frac{[1+h+h^2+h+14]-[1-6+14]}{h}

Can you break down what you did here? It's wrong, but it looks like it's wrong in quite a few places.

Determine equation of tangent line:

y = mx+b
9 = (-4)(-1) + b
9 = -4 + b
4 + 9 = b
b = 13

Therefore, the equation of the tangent to the curve is y= -4x + 13.

This looks good.

Completing the square:

x^2 - 6x + 14 = 0
x^2 - 6x + 9 + 14 - 9 = 0
(x-3)^2 + 14 - 9 = 0
(x-3)^2 + 5 = 0
(x-3)^2 = -5

<br /> \sqrt{(x-3)^2} = +- \sqrt{-5}<br />

x-3 = \sqrt{5}i
x = \sqrt{5}i + 3

and

x-3 = -\sqrt{5}i
x = -\sqrt{5}i + 3

What does finding the roots of the parabola have to do with finding the tangent line to the parabola at a point? And yes, you won't have real roots because the parabola is entirely above the x-axis.

 

[NoLimits]

I'm not sure the relevance (if any) of finding the roots. I saw it done in a video, and since the question wants a sketch of the graph, I assumed that I would be needing two values (x and y) for the vertex, hence the need for two roots. Sorry about the screw-up on the equation - I usually do a better job checking before I post.

This is the corrected line where I think it all went wrong:
<br /> 1+2h+h^2−6−6h+14<br />

 

[NoLimits]

So assuming I do not need the square roots, then the vertex must be (3, -5) and the y-intercept must be 9? Given the equations I have, it seems like they are the only possible values. However, when using a function graphing tool the y-intercept doesn't appear to be 9, nor does the vertex look like (3, -5). I can't seem to figure out how to go from the completed square to the values I need.

 

[Borek]

Have you read what the Mentallic wrote?

the parabola is entirely above the x-axis.

Completing the square is a way of finding roots of the equation - these are handy when you want to sketch the plot, but this equation has NO real roots, and doesn't cross the x-axis.

 

[NoLimits]

Have you read what the Mentallic wrote?
Completing the square is a way of finding roots of the equation - these are handy when you want to sketch the plot, but this equation has NO real roots, and doesn't cross the x-axis.

Yes I have read it, which is why I responded with what I did, though I did not completely understand what was meant by it. The question wants me to sketch it, so if the equation has no real roots and doesn't cross the x-axis, how is that possible? Just plug in x-values to the original equation?

 

[NoLimits]

Guess so. Thanks.

 

[Mentallic]

So assuming I do not need the square roots, then the vertex must be (3, -5)

Not exactly. As you've shown, after completing the square you end up with

y=(x-3)^2+5

Let's analyse this result for a second. We know that n^2\geq 0 for any real value of n, and most importantly, it's equal to 0 when n=0. So if we extend this to having (something)² then that's also always going to be greater than or equal to 0. It's again equal to 0 when something = 0.
What we have is (x-3)^2. This means when x-3 = 0, hence x=3, we get (x-3)^2=0 so then

y=0+5

So our minimum point on the parabola which is the vertex is at (3,5) which is above the x-axis and since this parabola has a positive coefficient of x², this means that it opens upwards, and this makes sense with why you were finding complex roots because it never cuts the x-axis.

and the y-intercept must be 9? Given the equations I have, it seems like they are the only possible values. However, when using a function graphing tool the y-intercept doesn't appear to be 9, nor does the vertex look like (3, -5). I can't seem to figure out how to go from the completed square to the values I need.

The y-intercept happens when you plug x=0 into the equation of the parabola. Similarly when you're finding roots of the equation, the x-intercept happens when you find y=0.