A parabola with vertex (h, k) has equation of the form:
y = a(x - h)² + k
→ vertex (k, h) = (3, 5), and a point on it is (-1, 6)
→ 6 = a(-1 - 3)² + 5
→ 6 = a(-4)² + 5
→ 1 = 16a
→ a = 1/16
→ The coefficient of the x² term is 1/16
The vertex of this parabola is at 5 5 When the x-value is 6 the y-value is -1. The coefficient of the squared expression in the parabola's equation is -6.
you didn't put any equations, but the answer probably begins with y= (x-4)^2+1
There are two forms in which a quadratic equation can be written: general form, which is ax2 + bx + c, and standard form, which is a(x - q)2 + p. In standard form, the vertex is (q, p). So to find the vertex, simply convert general form into standard form.The formula often used to convert between these two forms is:ax2 + bx + c = a(x + b/2a)2 + c - b2/4aSubstitute the variables:-2x2 + 12x - 13 = -2(x + 12/-4)2 -13 + 122/-8-2x2 + 12x - 13 = -2(x - 3)2 + 5Since the co-ordinates of the vertex are equal to (q, p), the vertex of the parabola defined by the equation y = -2x2 + 12x - 13 is located at point (3, 5)
The given equation is not that of a parabola since there are no powers of 2. Unfortunately, limitations of the browser used by Answers.com means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. Please resubmit your question spelling out the symbols as "plus", "minus", "equals" etc. And using ^ to indicate powers (eg x-squared = x^2).
You can work this out by taking the derivative of the equation, and solving for zero: y = x2 - 8x + 18 y' = 2x - 8 0 = 2x - 8 x = 4 So the vertex occurs where x is equal to 4. You can then plug that back into the original equation to get the y-coordinate: y = 42 - 8(4) + 18 y = 16 - 32 + 18 y = 2 So the vertex of the parabola occurs at the point (4, 2), leaving 2 as the answer to your question.
The vertex of this parabola is at -3 -1 When the y-value is 0 the x-value is 4. The coefficient of the squared term in the parabolas equation is 7
7
It is 1/16.
A coefficient is a number that accompanies a variable. For example, in the expression 2x + 4, the coefficient is 2.
To find the coefficient of the squared term in the parabola's equation, we can use the vertex form of a parabola, which is (y = a(x - h)^2 + k), where ((h, k)) is the vertex. Given the vertex at (3, 1), the equation starts as (y = a(x - 3)^2 + 1). Since the parabola passes through the point (4, 0), we can substitute these values into the equation: (0 = a(4 - 3)^2 + 1), resulting in (0 = a(1) + 1). Solving for (a), we find (a = -1). Thus, the coefficient of the squared term is (-1).
To find the coefficient of the squared expression in the parabola's equation, we can use the vertex form of a parabola, which is ( y = a(x - h)^2 + k ), where ((h, k)) is the vertex. Given the vertex is ((-4, -1)), the equation becomes ( y = a(x + 4)^2 - 1 ). When (y = 0) and (x = 2), substituting these values gives (0 = a(2 + 4)^2 - 1), leading to (0 = a(6^2) - 1) or (1 = 36a). Therefore, (a = \frac{1}{36}), which is the coefficient of the squared expression.
The vertex of this parabola is at -2 -3 When the y-value is -2 the x-value is -5. The coefficient of the squared term in the parabola's equation is -3.
The vertex of this parabola is at 5 5 When the x-value is 6 the y-value is -1. The coefficient of the squared expression in the parabola's equation is -6.
Vertex = (3, - 2)Put in vertex form.(X - 3)2 + 2X2 - 6X + 9 + 2 = 0X2 - 6X + 11 = 0=============The coefficeint of the squared term is 1. My TI-84 confirms the (4, 3) intercept of the parabola and the 11 Y intercept shown by the function.
-3
-5
-3