Feb 18, 2016 · How do you express 2x 4x2 + 12x + 9 in partial fractions?

Jun 4, 2018 · given a quadratic in standard form ; ax2 + bx + c then the x-coordinate of the vertex is ∙ xxvertex = − b 2a −2x2 +12x +9 is in standard form with a = −2,b = 12 and c = 9 xvertex = − 12 −4 = 3 substitute this value into the equation for y yvertex = − 2(3)2 +12(3) +9 = 27 vertex = (3,27)

Jul 22, 2015 · Re-write the given equation in "vertex form" to get: color (white) ("XXXX")vertex at (2,3) The general "vertex form" for a parabola is color (white) ("XXXX")y=m (x-a ...

Jun 17, 2016 · 3x^2+12x+9 =3 (x^2+4x+3) Use { cross product method} 1 3 1 1 x^2+4x+3= (x+1) (x+3) So 3 (x+1) (x+3)#

Sep 14, 2015 · color (blue) (x=3/2 4x^2–12x+9=0 The equation is of the form color (blue) (ax^2+bx+c=0 where: a=4, b=-12, c=9 The Discriminant is given by: color (blue) (Delta=b^2-4*a*c = (-12)^2- (4*4*9) = 144-144=0 As Delta=0 there is only one solution. The solution is found using the formula x= (-b+-sqrtDelta)/ (2*a) x = (- (-12)+-sqrt (0))/ (2*4) = (12+0)/8 = 12/8 = 3/2 The solution is color (blue) (x=3/2

Jun 17, 2017 · where ( h , k ) are the coordinates of the vertex and a is a constant. for a parabola in standard form y = ax2 +bx + c the x-coordinate of the vertex is xvertex = − b 2a y = 4x2 −12x + 9 is in standard form with a = 4,b = − 12,c = 9 ⇒ xvertex = − −12 8 = 3 2 substitute this value into function for y-coordinate

Apr 24, 2017 · rearrange terms and equate to zero ⇒ 3x2 −12x −9 = 0 divide all terms by 3 ⇒ x2 −4x −3 = 0 compare to standard form ax2 +bx + c = 0 here a = 1,b = − 4 and c = − 3 using the quadratic formula

May 7, 2016 · vertex: maximum point (2,3) Y-intercept = -9 X-intercept = 1 and 3 The given equation is -3x^2 +12x - 9 = y. Given that the coefficient (constant number in front) of x^2 is negative 3, for any NEGATIVE coefficient, the graph is a SAD smiley (has a maximum point).

Apr 21, 2016 · y=x^2+12x+9 =>y=x^2+12x+6^2-36+9 =>y= (x+6)^2-27 =>y+27= (x+6)^2 putting y+27 =Y and x+6=X we have Y=X^2 =>4xx1/4Y=X^2 Vertex of this Equation is (0,0) So the actual vertex putting X=9 and Y=0 x=-6 and y=-27 graph {x^2+12*x+9 [-58.53, 58.57, -29.24, 29.27]}

Feb 2, 2017 · x = -3+- (3sqrt (2))/2 Complete the square then use the difference of squares identity: a^2-b^2 = (a-b) (a+b) with a=2x+6 and b=3sqrt (2) as follows: 0 = 2 (2x^2+12x ...