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The curve is above this vertex and is symmetrical about its axis #x=-3#. This is parallel to y-axis. Put y= o.and solve #x^2+6x+5=0#@How do you use the important points to sketch the graph of y = x2 â 6x + 1? Algebra Quadratic Equations and Functions Quadratic Functions and Their Graphs*See the proof below We need cos2x=cos^2x-sin^2x cos^2x+sin^2x=1 (a-b)^3=a^3-3a^2b+3ab^2-b^3 Therefore, LHS=cos^3 2x+3cosx = (cos^2x-sin^2x)^3+3 (cos^2x-sin^2x) =cos *^See a solution process below: Step 1) Solve the first equation for y: 5x + y = -7 -color(red)(5x) + 5x + y = -color(red)(5x) - 7 0 + y = -5x - 7 y = -5x - 7 Step 2 *$-sqrt (9-6x-x^2)-3sin^-1 ( (x+3)/ (3sqrt2))+C Complete the square in the denominator: I=intx/sqrt (- (x^2+6x)+9)dx=intx/sqrt (- (x^2+6x+9)+9+9)dx I=intx/sqrt (18- (x+ *|The factored version is (x+3)^2 Heres how I approached it: I can see that x is in the first two terms of the quadratic, so when I factor it down it looks like: (x+a)(x+b) And when*Please see the process steps below; From how the question was understood, the interpretation is; We are looking for the unknown number, and let x be that number.. Statement! First