Decide the smallest distance from a point on the x-axis to a point on the curve y=2x^2+20x+56. If I draw the function I can calculate the distance, however can you calculate the function without drawing it? The problem is that the curve does not cut through the a-axis and I can therefore not make the function equal zero and use the quadric formula. Thank you in advance!
A function which does not cut the x-axis
- Thread starter Sara33
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Are there specific methods recommended?
Completing the square might do it.
2x^2 + 20x + 56 =
2(x^2 + 10x + _____) + 56 - 2*______ =
10/2 = 5
5^2 = 25
2(x^2 + 10x + 25) + 56 - 2*25 =
2(x+5)^2 + 6
What's the least possible value of (x+5)^2? Remember that it can't be negative!
Completing the square might do it.
2x^2 + 20x + 56 =
2(x^2 + 10x + _____) + 56 - 2*______ =
10/2 = 5
5^2 = 25
2(x^2 + 10x + 25) + 56 - 2*25 =
2(x+5)^2 + 6
What's the least possible value of (x+5)^2? Remember that it can't be negative!
J
JeffM
Guest
The hints I am going to give make sense only if you have studied parabolas. If you have not, there is no point in reading further.Sara33 said:
The graph of your equation forms a parabola in the x-y plane, right?
But it forms a very special parabola in that its axis of symmetry is perfectly vertical. Did you know that?
There is a special characteristic of the point of intersection of a parabola and a perfectly vertical axis of symmetry. Do you see what it is? A crude sketch may help you understand the general principle conceptually, but it is not necessary for solving your specific problem numerically.
In essence, what tkhunny has done is to give you an algebraic method to find the axis of symmetry for your equation.
Sara33 said:
\(\displaystyle \text{Let "v" in the position as a lower script stand for "vertex."}\)
\(\displaystyle Then \ (x_v, \ y_v) \ \ are \ \ the \ \ coordinates \ \ of \ \ the \ \ vertex.\)
\(\displaystyle \text{The value of the y-coordinate is your goal, because it is the distance from the vertex,}\)
\(\displaystyle \text{lowest point on the parabola (which here opens up), to the x-axis. As you mentioned,}\)
\(\displaystyle \text{this curve does not intersect the x-axis.}\)
\(\displaystyle x_v \ = \ \frac{-b}{2a}, \ \ where \ a = \ 2 \ \ and \ \ b = \ 20.\)
\(\displaystyle Calculate \ x_v. \ \ Then \ plug \ it \ back \ into \ y \ = \ x^2 + 20x + 56 \ to \ determine \ y_v.\)
\(\displaystyle Then \ y_v \ is \ the \ smallest \ distance \ you \ are \ seeking.\)
JeffM said:
I understand that the function forms a parabola. What is an axis of symmetry which is perfectly vertical? Does it mean that the axis of symmetry has an equal distance to both sides?
Thanks for everybodys help! I understand how to solve the exercise now.
J
JeffM
Guest
Because you have an answer to your problem, you may not bother to look at this thread again so I am going to keep this answer fairly brief. If you want to continue the conversation, let me know by responding to this post.Sara33 said:
Every parabola has an axis of symmetry, which is a straight line and is unique in that it is the only straight line to intersect the parabola in only 1 point. The axis of symmetry is called that because any line drawn perpindicular to it will intersect the parabola in two points that are EQUALLY distant from the axis.
If the equation of the parabola is in the form ax[sup:z2b1gm7m]2[/sup:z2b1gm7m] + bx + c, the axis of symmetry is a vertical line, and its equation will be x = [(- b) / 2a]. The unique point where a perfectly vertical axis intersects the parabola is the minimum (or maximum if the parabola opens down) point of the parabola. But of course a perfectly vertical axis will intersect the x-axis directly below that minimum point (or directly above the maximum). So, as lookagain explained, the formula for the distance between the point and the x-axis is just the absolute value of y.
J
JeffM
Guest
Well that is admirable indeed. GOOD FOR YOU.Sara33 said:
Let's approach this this way. The really important thing to learn from this problem is that there are frequently a number of different ways to solve a problem.
I intruded on tkhunny's instruction. As he said, you do not need to know anything about parabolas to solve this problem. There are in fact several ways to solve it. Work through his method, which involves a property of the quadratic. The great advantage of his method is that it utilizes what is probably the fundamental technique for working with quadratics, namely completing the square. Moreover, his method provides a hint of what is studied in differential calculus. Finally, you already know about completing the square.
My parabola method is just another path to the same end. It certainly is not the shortest path to solving this particular problem if you have to spend hours learning about parabolas before it makes sense. Don't worry: fairly soon, you will study conic sections and become quite bored with parabolas. Finish up with tkhunny (you made an error in your last response to him), and forget about parabolas temporarily. They will wait for you.
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Sara33 said:
Why would it be -5? There is a reason why completing the square was helpful. It makes these questions very easy.
2(x+5)^2 + 6
What is the minimum value of the left-hand term? 0 It can't be negative.
For what value does it take on zero (0). x = -5 Only inside the parentheses is in the consideration.
What is the minimum value fo the entire expression: 6 The left-hand term is zero at the minimum.
No need to do everything the hardest way possible, but unique answers don't care how you find them.