2016年7月14日 · The factors are (x+3) and (x+1) We are asked for factorise: x^2+4x+3 First notice that the function is a quadratic and so will have two factors. Since the coefficient of x^2 is 1, the factors will be of the form: (x+a)(x+b) We will assume that a and b are integers. Hence, we need to find a and b such that the product of the factors is equal to the given quadratic …

2016年6月4日 · x=1, 3 you start by finding a number that multiplies to 3 and adds to -4. they will both be negative as the product is positive and the sum negative. your numbers are -1 and -3. split up the 4x so that you can factor the equation further. x^2-4x+3 x^2-x-3x+3 x(x-1)-3(x-1) (x-3)(x-1) x=3 or 1

2017年10月15日 · x^2+4x+3 is already in standard polynomial form. Standard form x^2+4x+3 is already in standard form for a quadratic polynomial. Standard form of a polynomial is a sum of terms in descending order of the power of the variable. Having the highest power of x first reflects the fact that this term is dominant for larger values of x. If the coefficient of x^2 is positive, then …

2017年7月22日 · How do you solve #x^2+4x+3=0# using the quadratic formula? Algebra Quadratic Equations and Functions Quadratic Formula. 1 Answer

2017年12月16日 · See below Roots (intercepts) Let's first solve this parabola to obtain the roots. Factorise x^2-4x+3: (x-3)(x-1) x-3=0, x-1=0 x=3, x=1 Therefore, the roots are x=3 and x=1 Vertex The vertex can be found by using the formula -b/(2a). In our case, a=1, b=-4 Substitute a=1, b= -4 into the formula: -(-4)/((2)(1)) -(-2) 2 Therefore, the x coordinate of the vertex is at x=2 *Note …

2017年10月17日 · The general vertex form is #color(white)("XXX")y=(x-color(red)a)^2+color(blue)b# with vertex at #(color(red)a,color(blue)b#)

2017年2月28日 · How do you solve quadratic inequality, graph, and write in interval notation #x^2 - 8x + 15 >0

2015年7月29日 · color(red)( f(x) = (x-2)^2-1) > The vertex form of a quadratic is given by y = a(x – h)^2 + k, where (h, k) is the vertex. The "a" in the vertex form is the same "a" as in y = ax^2 + bx + c. Your equation is f(x) = x^2-4x+3 We convert to the "vertex form" by completing the square. Step 1. Move the constant to the other side. f(x)-3 = x^2-4x Step 2. Square the coefficient of x and …

2015年4月8日 · Answer: => y <= 1 Solution: The range of this function is the set of the values of y that the f can take, So, letting y= f(x) y = -x^2 + 4x - 3 => x^2 - 4x + 3 +y = 0 For real x , b^2 - 4ac >=0 For the quadratic ax^2 + bx + c = 0 Applying this to our equation, => (-4)^2 - 4(1)(3 + y) >= 0 => 16 - 12 - 4y>= 0 => 4 >= 4y => y <= 1 In other terms (-oo, 1]

2015年8月11日 · The curve is concave upwards. It is a quadratic function. It has one turning point. Your graph must include that. x^2 has a positive sign. The curve is concave upwards. It has a minimum. x = (-b)/(2a) = (-4)/(2 xx 1)= -2 At x = -2 the curve turns. Take a few points on either side of '-2'. -5, -4, -3, -2, -1, 0, 1 - This is the range of values. Find the corresponding y co …