وبلاگ خانه تبدیل شباهنگ مرکز تخصصی تبدیل صدا و تصویر
To convert from f(x) = ax2 + bx + c shape to vertex shape: Method 1: Complete square To convert a square shape of y = ax2 + bx + c to a vertex shape, y = a(x – h)2+ k, use the process of completing the square. Let`s take an example. Before I begin, I realize that a = 1. Therefore, I can immediately apply the “Complete square” steps. If written in “vertex shape”: • (h, k) is the vertex of the parabola, and x = h is the axis of symmetry. • The h represents a horizontal offset (how far to the left or right the graph moved by x = 0). This is the vertex form of the square function, where left( {h,k} right) is the vertex or “center” of the square function or parabola. In a regular algebra class, completing the square is a very useful tool or method for converting the quadratic equation of the form y = a{x^2} + bx + c, also known as the “standard form”, into the form y = a{(x – h)^2} + k, known as the vertex form. Method 2: Use the “sneaky processing” seen above to convert to vertex shape: • Note that the h-value is subtracted in this form and the k-value is added. If the equation is y = 2(x – 1)2 + 5, the value of h is 1 and k is 5.
If the equation is y = 3(x + 4)2 – 6, the value of h is -4 and k is -6. STEP 1: Identify the coefficient of the linear term of the quadratic function. This is the number added to the term x. Convert y = 2×2 – 4x + 5 to vertex shape and specify the vertex. This quadratic equation has the form y = a{x^2} + bx + c. However, I have to rewrite it with a few algebraic steps to make it look like this. Example 1: Find the vertex shape of the square function below. • The k represents a vertical offset (how far the graph has moved up or down with respect to y = 0). . . . Think about it: if I add 4 on the right side of the equation, then I technically change the original meaning of the equation.
So, to leave it unchanged, I have to subtract the same value I added on the same page of the equation. STEP 4: Now I take the output large{9 over 4} and paste it in parentheses. . y = 2(x – 1)2 + 3 y = 2(x2 – 2x + 1) + 3 y = 2×2 – 4x + 2 + 3 y = 2×2 – 4x + 5. STEP 4: Now, tap on the trinom in parentheses as a square of a binomial and simplify the external constants. For this problem, we chose (to the left of the axis of symmetry): Convert the vertex to y = ax2 + bx + c Shape: STEP 2: I take this number, divide it by 2 and place it (or raise it to power 2). STEP 3: The output of step #2 is added up and subtracted from the same side of the equation to keep it in balance. STEP 1: Factor 2 only to the terms with the variable x. The approach to this problem is slightly different because the value of “a” is not equal to 1, a ne 1. The first step is to factor the coefficient 2 between terms only with x variables. STEP 5: Since I added large{9 over 2} to the equation, I have to subtract the whole equation with large{9 over 2}, also to compensate for this.
STEP 2: Identify the coefficient of the term x or linear term. .
