7. Vertex and tangent 6 Example 1. At the right we have drawn the graph of the cubic polynomial 4 f(x) = x2 (3 – x). Notice how the structure of the graph matches the form of the algebraic expression. The zeros are at x=0 and x=3, with a double zero at x=0 so that the graph doesn’t cross the axis at the origin. Our interest here it in the “vertex” which is the maximum point on the “hump” between 0 and 3. In the diagram it seems to be at the point (2,4). But is it? Exactly? Might it be at (2.01, 3.99)? Your job is to settle this question algebraically. Show that the vertex is exactly at (2,4). y 2 x 0 -1 0 1 2 3 4 -2 -4 Show that the vertex is exactly at the point (2,4). Solution. First of all, we note that the point (2,4) is indeed on the graph. If we put x=2 into the formula, we get f(2) = (2)2 (3 – 2) = 4 This is the problem we throw out to the class. and that shows that the height of the graph at x=2 is exactly 4. So to solve the problem, all we have to show is that the graph never gets higher than y=4 on the interval [0,3]. That is, we have to show that x2 (3 – x) < 4 for 0≤x≤3, x≠2. How do we do that? A useful step is to put everything over on the left-hand side and leave 0 on the right: x2 (3 – x) – 4 < 0 for 0≤x≤3, x≠2. – x3 + 3x2 – 4 < 0 for 0≤x≤3, x≠2. Expand: Now where are we? We are required to show that a certain polynomial is negative for certain values of x. How do we do that? 7. vertex and tangent Often in mathematics our problem reduces to demonstrating a certain inequality: A < B. Very often it turns out that the best strategy is to rewrite this as A – B < 0. Simple though it is, this transformation can make a great difference. Zero is a nice number to have as part of the action. 1 Finding the sign. We have to show that – x3 + 3x2 – 4 < 0 for 0≤x≤3, x≠2. Thus we have to show something about the sign of the polynomial. Now to analyze the sign of a polynomial, the best thing is to have it in factored form. So we want to factor – x3 + 3x2 – 4. How do we do that? Well, it’s a cubic polynomial so we can factor it if we can find a “first zero”? Can we find a value of x for which – x3 + 3x2 – 4 = 0 ? That's a key question, and the answer is yes we can!––the context of the problem tells us that x=2 has to satisfy this equation. Indeed the above equation reads f(x) – 4 = 0 and we showed at the beginning that f(2) – 4 = 0. So the factor theorem tells us that (x–2) is a factor of – x3 + 3x2 – 4. Divide and find the quotient: This is a beautiful application of the Factor Theorem. Note that the “first zero” came out of the context of the problem. That happens quite often. – x3 + 3x2 – 4 = (x–2)(–x2 + x + 2) = – (x–2)(x2 – x – 2) Note that for ease of handling, we have taken a minus sign out of the quadratic. Now factor this quadratic. (x2 – x – 2) = (x–2)(x+1). Putting it all together: – x3 + 3x2 – 4 = – (x–2) (x–2) (x+1) = – (x–2)2 (x+1) and we have completed the factorization. Now what do we have to show? We have to show that this is negative between 0 and 3, except at x=2. But this is now obvious, since (x+1) is positive on this interval, and (x–2)2 is positive except at x=2. And the proof is complete. 6 Note. One of our students, by working a bit longer with the picture, found a rather more effective route through the beginning analysis. Once she verified that the point (2,4) was on the curve, she said, “Let’s drop the curve by 4 units. Then what we can show is that the new curve lies below the x–axis (for x positive), except of course at x=2.” y 4 2 x 0 -1 0 1 2 3 -2 -4 7. vertex and tangent 2 4 Example 2. At the right we have drawn the graphs of the 4th degree polynomial yP = x2 (9 – x2) 30 and of the line 20 yL = 12 + 4x. Show that the line is tangent to the curve at x=2. [That is, show that the curve contacts the line, but stays on one side of it, at least "locally."] 10 0 -4 Solution. I guess the first thing to note is that the curve does contact the line at x=2. Indeed, yP(2) = 22 (9 – 22) = 20 yL(2) = 12 + 4(2) = 20 so at that point they both have height 20. And for other values of x, the line appears to be above the curve, at least for positive x. Though the resolution of the diagram certainly does not allow us to be certain about this. Anyway, that's what we want to verify, that the line is above the curve for x≠2, at least for x>0. Our strategy will be focus on the amount that the line is above the curve, and to see how that changes with x. That's just the distance from the line down to the curve. Algebraically it's the difference between the two y-expressions: -3 -2 -1 0 1 2 3 -10 -20 This might strike you as an unreasonably difficult problem for grade 11, but in fact it's quite wonderful. The steps and ideas are all simple applications of standard material, for example the factor theorem, ideas about change of sign and the conditions which ensure that one graph is above another. But when everything is put together in the right order, which requires clear thought and organization, we get a sophisticated result. yL – yP = (12 + 4x) – x2(9 – x2) = x4 – 9x2 + 4x + 12. We want to show that this is positive for x>0, x≠2. And of course the way to do that is to write it in factored form. Now how do we do that? Well we need to find some zeros. Okay––what are the zeros of this polynomial? Well they are the zeros of yL – yP. Those are the points at which yL and yP are equal. And those are the points at which the line and the curve intersect. Well we've just verified that x=2 is such a point. So that gives us a factor of (x–2). Can we find other factors? 7. vertex and tangent 3 4 Finding the factors. It appears from the picture that x=–3 might also be a zero of yL – yP . Indeed both functions seem to take the value zero at that point, and this is easy to check from the algebraic expressions: 30 yP(–3) = (–3)2 (9 – (–3)2) = 0 10 20 yL(–3) = 12 + 4(–3) = 0 So that gives us a factor of (x+3). And there seems to be an intersection near x=–1. Let's investigate: 2 0 -4 -3 -2 -1 0 1 2 3 -10 2 yP(–1) = (–1) (9 – (–1) ) = 8 -20 yL(–1) = 12 + 4(–1) = 8 and they intersect. So that gives us a factor of (x+1). Those are the three intersection from the picture, so we write: yL – yP = x4 –9x2 + 4x + 12 = (x–2) (x+3) (x+1)(x+k). We have put one unknown factor (x+k) on the end because the polynomial is of degree 4 and three linear factors would only make degree 3. How can we find that last factor? We need only compare the constant term of both sides: Alternatively, once we have the two zeros x=2 and x=–3, we could just divide the whole polynomial by (x–2)(x+3) and we'd get a quadratic which we could then factor. Try it. 12 = (–2) (+3) (+1)(+k) we can see that k has to equal –2 so the remaining factor has to be (x–2). That's a second appearance of (x–2). Interesting. So our candidate for the factorization is yL – yP = x4 –9x2 + 4x + 12 = (x–2)2 (x+3) (x+1). To check that this holds, we simply expand the right-hand side, and this is best done in two steps (Problem 1): (x–2)2 (x+3) (x+1) = (x2–4x+4) (x2+4x+3) = x4 –9x2 + 4x + 12 and the factorization is correct. Now we can easily analyze the sign of the factored expression. For all x>0, every term in the factorization is positive. Well, there’s one exception––the term (x–2)2 is positive except at the one point x=2. So the entire expression yL–yP is positive except at x=2, and that tells us that the line is above the curve for positive x, and is tangent to it at x=2. 7. vertex and tangent 4 4 30 Example 3. At the right we have drawn the graph of the 4th degree polynomial yP = x2 (x2 – 4) 20 and of the line yL = 16(x–2). 10 Show that the line is tangent to the curve at x=2. 0 Solution. The curve and the line clearly intersect at x=2 as they both cross the x-axis at this point. For other values of x, the curve appears to be above the line. That's what we want to verify, and as before, our strategy will be focus on the separation between the two graphs. Algebraically that's the difference between the two y-expressions: -3 -2 -1 0 1 2 3 -10 This is a rather more demanding example, but I find its beauty irresistible. yP – yL = x2 (x2 – 4) – 16(x–2) = x4 – 4x2 – 16x + 32. We want to show that this is positive for x≠2. As before, the way to do that is to write the expression in factored form, and the key step here is to identify the zeros. These are the points at which the line and the curve intersect, and we have x=2 as such a point. In fact, if the curve indeed does not cross the line at this point, x=2 will have to be a double zero, and that would give us a factor of (x–2)2. As in Example 1, the way to see this is to think of the graph of yP–yL against x. It hits the x-axis at x=2, but doesn't cross it. Are there other intersections? Well if there are they don't appear in the diagram, and the way the curve is curving away, it doesn't look as if there ever will be. What to do? Well let's try to divide out the factor (x–2)2 and see what we get: x4 – 4x2 – 16x + 32 = (x–2)2 ( ??? ) To find the last term we first expand the square. x4 – 4x2 – 16x + 32 = (x2–4x+4) ( ??? ) We could use long division to find what goes in the last bracket, but we prefer to write it down by inspection. Look at the expression on the left: the x4 at the beginning tells us that the bracket begins with x2, and the +32 at the end tells us that the bracket ends with +8. So we need only work out the xterm in the bracket: x4 – 4x2 – 16x + 32 = (x2–4x+4) (x2 + ?x +8) From the fact that the expression on the left has no x3-term, we deduce that the x-term in the bracket must be +4x. We get: x4 – 4x2 – 16x + 32 = (x2–4x+4) (x2+4x+8) and we can now check that this holds by expanding the right hand side. (Problem 1). 7. vertex and tangent 5 Let’s not lose sight of what we’re trying to get––and that’s the complete factorization of the 4th degree polynomial on the left. So far we have: x4 – 4x2 – 16x + 32 = (x–2)2 (x2+4x+8) and to finish things off we need only the factorization of the quadratic on the right: y = x2+4x+8. Nothing obvious presents itself, so we use the quadratic formula to find its roots. We get: x = − 4 ± 16 − 32 2 and the discriminant is negative!! So there are no zeros. So the 4th degree polynomial is already fully factored. What we want to know about is the sign of the whole expression. Well since the quadratic polynomial x2+4x+8 has no zeros, its graph can't cross the axis. What does that mean?–– that its graph must either live above the axis or below the axis, and therefore the expression must either always be positive or always be negative. 12 It's not hard to decide which of these is the case. The quadratic takes value 8 at x=0, so it must always be positive. [Alternatively, you can verify directly that the vertex of the parabola is at (–2,4) and it opens up, so it will clearly always be positive.] 10 8 6 4 We are finished. The distance between the curve and the line is: 2 yP – yL = x4 – 4x2 – 16x + 32 = (x–2)2 (x2+4x+8) The square term (x–2)2 is positive except at x=2 and the last quadratic is always positive, so the whole expression is positive except at x=2. And we are done––the curve is above the line except at x=2, and therefore the line is a tangent. 7. vertex and tangent 0 -5 -4 -3 -2 -1 0 1 graph of the parabola x2+4x+8. 6 Problems y 30 1. At the right we have drawn the graph of the degree-4 polynomial 20 y = x3 (4 – x). 10 Our interest here it in the “vertex” which is the maximum point on the “hump” between 0 and 4. In the diagram it seems it might be at the point (3,27). But is it? Your job is to settle this question algebraically. Show that the vertex is exactly at (3,27). x 0 -2 -1 0 1 2 3 4 5 -10 -20 2. Let f(x) = x3 – 6x2 + 9x – 2. Verify that f(2) = 0, and use this fact to find all three zeros of the polynomial and sketch a reasonable graph of f(x). Now use the methods of this section to show that the graph has vertices at x=1 (a max point) and at x=3 (a min point). 20 3. At the right we have drawn the graph of the degree-4 polynomial y = x2 (8 – x2). Our interest here it in the “vertices” which are the maximum points on the two “humps”. In the diagram these seem to be at x = ±2. But are they? Your job is to settle this question algebraically. 4. You already know how to find the vertex of the parabola y 15 10 5 x 0 -3 -2 -1 -5 0 1 2 -10 y = x (8 – x). Verify using the methods of this section that this is the highest point on the curve. 5. Observe that f(x) = x4 – 8x2 + 20 has value 4 at x = ±2. Verify that this is the minimum value attained by f(x), that is, that f(x) ≥ 4 for all x. 7. vertex and tangent 7 3 6. Check that the factorizations of Examples 2 and 3 both hold by expanding the right hand side. In each case show clearly the intermediate step in which the 9 different terms of the expansion are displayed. Example 2: x4 –9x2 + 4x + 12 = (x2–4x+4) (x2+4x+3) Example 3: x4 – 4x2 – 16x + 12 = (x2–4x+4) (x2+4x+3) . 7. Find a quadratic polynomial yP = ax2+bx+c which is tangent to the line yL = 2(x–1) at the point x=2. [Hint: First draw a rough picture of the line and the parabola so you can see what you're heading for. Then think about the graph of the difference z = yP – yL , and attempt to find an equation for it. From that you can find yP as z+yL .] 8. At the right we have drawn the graph of the cubic polynomial y = x (9 – x2). 20 15 And of the line 10 y = 16 – 3x. Show that the line is tangent to the curve at x=2. [That is, show that the curve contacts the line, but stays on one side of it, at least "locally."] 5 0 -4 -2 0 2 4 -5 -10 9. Show that the line 3x + 4y = 25 -15 is tangent to the circle x2 + y2 = 25. Start by drawing a picture. Find the point where they intersect. (a) Following the methods of this section. Solve the equations of both the line and circle for y in terms of x, and do what you have to do to show that the line is always above the circle except at the one point.] (b) Using geometry. Show that the line is perpendicular to the radius of the circle drawn to the point of contact. 7. vertex and tangent 8 6 10. Draw a rough sketch of the parabola In Problems 10-12 we ask you to draw rough sketches. How are you supposed to know how to handle x3 and x4 ? But there are a number of features you can use as a guide. Where it's positive or negative or zero, where it increases or decreases, bilateral symmetry, etc. 2 yP = x . Show that the tangent line to the curve at x=1 has slope 2. [That is, show that the line of slope 2 which intersects the parabola at x=1 is tangent to the curve.] 11. Draw a rough sketch of the graph of the cubic polynomial yP = x3. Show that the tangent line to the curve at x=2 has slope 12. [That is, show that the line of slope 12 which intersects the curve at x=2 is tangent to the curve.] 12. Draw a rough sketch of the graph of the 4th degree polynomial yP = x4. Show that the tangent line to the curve at x=1 has slope 4. [That is, show that the line of slope 4 which intersects the curve at x=1 is tangent to the curve.] 30 20 13. In Example 2, we considered the function yL – yP which at any x was the vertical distance between the line and the curve, where this distance is taken as positive if the line is above the curve and negative otherwise. Draw the graph of this function and use it to explain why it has x=2 as a double zero. 10 0 -4 -3 -2 -1 0 1 2 3 4 -10 -20 15 14. At the right we have drawn the graph of the 4th degree polynomial yP = x3 (x+2). 10 Show that the tangent line to the curve at x=1 has slope 10. [That is, show that the line of slope 10 which intersects the curve at x=1 is tangent to the curve.] 5 0 -3 -2 -1 0 1 2 -5 7. vertex and tangent 9

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