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Topic: **How to write a math expressions****Question:**
Hi there, I need to find a parabola and have been given three tangents of that parabola:
y=0
y=(-6/7)x-2
y=(4/5)x-2
If anyone could help that would be greatly appreciated (p.s. this is for year 12 maths, so calculus okay)

June 25, 2019 / By Murtagh

Let's take each of the tangent lines to be written at x1, x2 and x3, which are to be determined. For a function f(x), its tangent line at xi is given as y = f'(xi)*(x-xi) + f(xi). The first tangent line is then y = 0 = 0*(x-x1) + 0, But if the parabola is given by f(x) = a*x^2+b*x+c, this tangent line tells us that f(x1) = f'(x1) = 0, from which b and c can be determined in terms of a. Thus, f(x) = a*(x-x1)^2, and so we now need to figure out a and x1. This can be done by re-writing the second line as y = -6/7x-2 =-6/7*(x-x2) -2-6/7*x2. From the third equation, y = 4/5x-2 =4/5*(x-x3) -2-4/5*x3. and by comparing this to f(x2), f(x3) f'(x2) and f'(x3), we get the four equations 2*a*(x2-x1) = -6/7 2*a*(x3-x1) = 4/5 a*(x2-x1)^2 = -6/7*x2-2 a*(x3-x1)^2 = 4/5*x3-2 Substituting the first and second equations into the third and fourth to eliminate a, we get -3/7*(x2-x1) = -6/7*x2-2 2/5*(x3-x1) = 4/5*x3-2 and thus x2 = -x1-14/3, x3 = -x1+5. Using these expressions to eliminate x2 and x3, we get 2*a*(-2*x1-14/3) = -6/7 2*a*(-2*x1+5) = 4/5 and subtracting these two equations, eliminates x1, so that (28/3+10)a = 6/7+4/5 ==> a = 3/35 Finally, we can use this to determine that x1 = 1/6, and so the parabola is f(x) = 3/35*(x-1/6)^2

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Dear Asker, I'm going to assume that solving quadratic equations is within bounds. (There are methods based on calculus, but since you said 10th grade, I'll assume that these are out of bounds for a while.) All non-vertical lines through (2,1) have the form y - 1 = m(x - 2). We're looking for values of the slope m for which the line will be tangent to the parabola. This means that the line will intersect the parabola exactly once. Thus, when we solve the system y - 1 = m(x - 2) y = x^2 we want just one solution. Substituting from the second equation into the first gives x^2 - 1 = m(x - 2). Rearranging this expression gives x^2 - mx + (2m - 1) = 0. This quadratic will have exactly one real root if its discriminant b^2 -4ac is zero. Thus, m^2 - 4*1*(2m - 1) = 0. The roots of this quadratic are 4 + 2*sqrt(3) and 4 - 2*sqrt(3). These are the slopes of the lines through (2,1) that can be tangent to the given parabola. The line you mentioned is not tangent to the parabola. If you have a graphing calculator, try graphing the parabola and line together, with the range on x restricted to -2 to 2. You'll notice that the line crosses the parabola and is not tangent to it. I hope this helps. bj ..

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(4,3) is on the graph of y your tangent is y-3 = (1/4)(x-4) because you evaluated your function at 4, and your first dirivative of your function at 4, that is f'(4), to get your tangent's slope at the point of tangency (4,3). Then you write the line equation knowing the slope of the line and a point on the line. - - - - - - - - y-4 = (-1/7)(x+2) is your normal because your first derivative of your function, evaluated at x= -2 gives 7, that is f'(-2) = 7; so the normal (perpendicular) has slope that is the negative reciprocal (of the tangent's slope). Again you have slope (of the normal line) and a point on the line from which to write the equation of the line. - - - - - - - - - k=-2 from taking the derivative of the function, "plugging in" 2 for x (in the derivative) and setting that expression = 4 in order to solve for k - - - - - - - - - - right of 1 y is concave up; left of 1 y is concave down because the second derivative of the function becomes zero at x=1, and is negative left of 1, and positive right of 1. - - - - - - - - (Last one later.) The function is increasing in the open interval (neg infinity, [6-sqrt6]/6), decreasing in the open interval ([6-sqrt6]/6, [6+sqrt6]/6), then increasing in the open interval ([6+sqrt6]/6, pos infinity) This is determined from looking at the sign behavior of the first derivative over the domain of definition of the function. The first derivative is "+" when the function is increasing, "-" when the function is decreasing. Your function has no absolute max or min; but has a relative max at x=[6-sqrt6]/6 and a relative min at x=[6+sqrt6]/6. At these critical points I refer to the function as neither increasing of decreasing; some texts might include them -- the critical points -- in the function's interval(s) of increase or decrease; I think that's ambiguous, but adapt to that constraint if it's required.

~find the equation of the tangent to y = (the squared root of x) + 1 at x = 4 Using the point slope formula: You first take the derivative of it and plug in your x to find the slope. Then you go back to ur original func. and plug in your x to find your y. Then plug in to make your equation y - y1=m (x - x1) Simplify and get your equation. *others might use different equation like slope intercept one. ~Find the equation of the normal to y = x ^ 3 - 5 x + 2 at x = -2 ok you do the same thing but you know have to change the sign (neg., pos.) of the slope and flip it (i.e. (1/2) change to (2/1)). ~Find k if the tangent to y = 2 x^3 + k x^2 - 3 at the point where x=2 has slope 4. Take Der. and plug in slope which is your y' and your x to find what k is maybe not sure about this one sorry. The other problems take too long i think I dont like them.

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