1. Find an equation for the straight lines which pass through the point (1, -2) and are tangent to the parabola with equation y = x2.

Would my two equations be y = x2 and -2 = m + c? Where do I go from here? Usually I would make equation 1 equal equation 2, but y is not equal to -2.

2. Solve the following for x by completing the square:

2x2 - 4x + 1 = 0

What I tried was:

2(x2 - (4/2)x + (1/2)) =0

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

(x - 1)2 - (1/2) = 0

x - 1 = sqrt(1/2) = 0

x = 1 + sqrt(1/2)

2. Re: Moar Quadratics!

Let's work the first problem before moving to the second. Using the point-slope formula, with unknown slope $m$, what is the family of lines which pass through the given point?

3. Re: Moar Quadratics!

y - y1 = m(x - x1)

y + 2 = m(x - 1)
y = mx - m + 2
y = m(x - 1) + 2

4. Re: Moar Quadratics!

Yes, this is correct:

$y-(-2)=m(x-1)$

and so:

$y+2=m(x-1)$

This is what you correctly found,

Now subtract 2 from both sides to solve for y...I think you just got a little careless when you did this.

5. Re: Moar Quadratics!

Yep, I see it.

y = m(x - 1) - 2

Would we then do something like...?

x2 = m(x - 1) - 2

-x2 + m(x - 1) - 2 = 0

If we were to solve this by equating the discriminant to 0, would the b constant of said discriminant be?

6. Re: Moar Quadratics!

First, you want to write the quadratic in standard form $ax^2+bx+c=0$. I recommend distributing the m, and then multiplying through by -1. What do you get?

7. Re: Moar Quadratics!

x2 - mx + m + 2 = 0

8. Re: Moar Quadratics!

Yes, so equate the discriminant to zero and solve for m...what do you find?

9. Re: Moar Quadratics!

What is the b value of the discriminant? Is it m?

10. Re: Moar Quadratics!

It is -m actually, but this won't affect the computation of the discriminant, but it is good to recognize that it we have:

$(1)x^2+(-m)x+(m+2)=0$

This way we can easily see that:

$a=1,\,b=-m,\,c=m+2$.

11. Re: Moar Quadratics!

Ah, I see. I'll complete this equation later and post my results.

12. Re: Moar Quadratics!

m2 - 4m - 8 = 0

Therefore m = 2 +- 2sqrt3.

So do we then just substitute this m value back into y = mx - m - 2?

13. Re: Moar Quadratics!

Yes, this will give you the two lines that satisfying the given conditions:

y=x^2,y=(2+2sqrt(3))(x-1)-2,y=(2-2sqrt(3))(x-1)-2 where x=-2 to 4 - Wolfram|Alpha

14. Re: Moar Quadratics!

Awesome.

Where did I go wrong with the following equation?

2(x2 - (4/2)x + (1/2)) =0

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

(x - 1)2 - (1/2) = 0

x - 1 = sqrt(1/2) = 0

x = 1 + sqrt(1/2)

15. Re: Moar Quadratics!

When you took the square root of both sides, you did not consider the negative root.

If $x^2=y$ then $x=\pm\sqrt{y}$

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