Vertex The vertex is easily determined by completing the square. y = 3(x^2 + 4x + p) + 15 p = (b/2)^2 p = (4/2)^2 p = 4 y = 3(x^2 + 4x + 4 - 4) + 15 y = 3(x^2 + 4x + 4) - 4(3) + 15 y = 3(x + …

2018年6月4日 · #int dx/sqrt(4x^2+12x-5) = 1/2 ln abs (2x+3+sqrt(4x^2+12x-5))+C# and by differentiating we can verify that the solution is valid also for #(2x+3)/sqrt14 < -1# . Answer link

2017年5月24日 · What are the points of inflection, if any, of #f(x) =12x^5+15x^4–240x^3+6#? Calculus Graphing with the Second Derivative Determining Points of Inflection for a Function

2017年8月28日 · The easiest way is to convert the given equation into vertex form to determine that the vertex is at #(-2,-7)#

-12x^5+2x^4-5x^3+3x^2-8x-6 I am assuming the 3 in the -5x term should be an exponent. 3x^2-8x-12x^5-5x^3+2x^4-6 To write the polynomial in standard form, write it in order of decreasing …

2018年1月15日 · just add and subtract so we have x^2 - 12x + 5 = 0 x^2 - 12x + 36 - 36 + 5 = 0 (x+6)^2 -31 = 0 therefore x + 6 = +-31^(1/2) so x = +-31^(1/2) - 6 hope u find it helpful :)

2015年9月13日 · It is (12x^5) /( 3x^7)=(3*4*x^5)/(3*x^5*x^2)=4/x^2

2016年11月15日 · Please see the explanation. The given equation is the equation of a parabola that opens upward (or downward). The vertex form of the equation of a parabola that opens …

2015年5月2日 · We can Split the Middle Term of this expression to factorise it In this technique, if we have to factorise an expression like ax^2 + bx + c, we need to think of 2 numbers such …

2015年7月14日 · 12x^5+6x^3+8x^2 = 2x^2(6x^3+3x+4) = 12x^2(x-x_1)(x-x_2)(x-x_3) where x_1, x_2, x_3 are defined below. 12x^5+6x^3+8x^2 = 2x^2(6x^3+3x+4) Let f(x) = 6x^3+3x+4. This …