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How do you find the vertex of an absolute value function?
There are three main types of vertices for an absolute value function. There are some vertices which are carried over from the function, and taking its absolute value makes no difference. For example, the vertex of the parabola y = 3*x^2 + 15 is not affected by taking absolute values. Then there are some vertices which are reflected in the x-axis because of the absolute value. For example, the vertex of the absolute value of y = 3*x^2 - 15, that is y = |3*x^2 - 15| will be the reflection of the vertex of the original. Finally there are points where the function is "bounced" off the x-axis. These points can be identified by solving for the roots of the original equation. -------------- The above answer considers the absolute value of a parabola. There is a simpler, more common function, y = lxl. In this form, the vertex is (0,0). A more general form is y = lx-hl +k, where y = lxl has been translated h units to the right and k units up. This function has its vertex at (h,k). Finally, for y = albx-hl + k, where the graph has been stretched vertically by a factor of a and compressed horizontally by a factor of b, the vertex will be at (h/b,ak). Of course, you can always find the vertex by graphing, especially since you might not remember the 2nd or 3rd parts above.
How do you find the axis of symmetry and vertex of y equals x squared plus 6x plus 10?
By completing the square y = (x+3)2+1 Axis of symmetry and vertex: x = -3 and (-3, 1) Note that the parabola has no x intercepts because the discriminant is less than zero
What is the vertex of the graph of the function y equals x2-4x plus 3?
To find the extreme value of the parabola y = x2 - 4x + 3 ...(1) Take the derivative of the equation.y = x2 - 4x + 3y' = 2x - 4(2) Set the derivative = 0 and solve for x.y' = 2x - 40 = 2x - 42x = 4x = 4/2x = 2(3) Plug this x value back into the original equation to find the associated y coordinate.x = 2y = x2 - 4x + 3y = (2)2 - 4(2) + 3y = 4 - 8 + 3y = -1So the vertex is at (2, -1).
How do you find the vertical asymptote for the logarithmic function f of x equals log x?
The only way I ever learned to find it was to think about it. The function f(x) = log(x) only exists of 'x' is positive. As 'x' gets smaller and smaller, the function asymptotically approaches the y-axis.
Find the measure of this angles m1 equals 123 m8 equals?
Find the measure of this angles m1 equals 123 m8 equals?
What is the vertex of the function y equals x2-4x 1?
To find the vertex of the quadratic function ( y = x^2 - 4x + 1 ), we can use the vertex formula ( x = -\frac{b}{2a} ), where ( a = 1 ) and ( b = -4 ). This gives ( x = -\frac{-4}{2 \cdot 1} = 2 ). Substituting ( x = 2 ) back into the equation, we find ( y = 2^2 - 4(2) + 1 = -1 ). Thus, the vertex of the function is at the point ( (2, -1) ).
What is the vertex form of a quadratic function and how do you find the vertex when a quadratic is in vertex form?
The vertex form of a quadratic function is expressed as ( f(x) = a(x-h)^2 + k ), where ( (h, k) ) represents the vertex of the parabola. To find the vertex when a quadratic is in vertex form, simply identify the values of ( h ) and ( k ) from the equation. The vertex is located at the point ( (h, k) ).
How do you find the vertex of the parabola y equals -4x2 - 16x - 11?
You would convert it to vertex form by completing the square. You can also find the optimum value as optimum value and vertex are the same.
Find the vertex of y equals x2 plus 10x plus 22?
-2-5
Find the vertex and equation of the directri for y2 equals -32x?
y2 = 32x y = ±√32x the vertex is (0, 0) and the axis of symmetry is x-axis or y = 0
How do you write y equals x minus 4 x plus 2 in vertex form and find the vertex?
The given equation is y = x - 4x + 2 which can be written as y = -3x + 2 This is an equation of a straight line. Therefore it has no vertex and so cannot be written in vertex form.
Do you have find the croodinates for the vertex for the quadratic function?
Yes, the coordinates for the vertex of a quadratic function in the form (y = ax^2 + bx + c) can be found using the formula (x = -\frac{b}{2a}) to determine the x-coordinate. Once you have the x-coordinate, substitute it back into the original equation to find the corresponding y-coordinate. This gives you the vertex in the form ((x, y)).
how to find the vertex angle?
The vertex angle is connected to the vertex point
Find the x-coordinate of the vertex y equals x exponent 2 plus 12x plus 21?
y = x2 + 12x + 21At the max or min point, the first derivative of the function = 0.2x + 12 = 02x = -12x = -6
What is the vertex of the quadratic function f(x)x2 plus c?
The vertex of the quadratic function ( f(x) = ax^2 + bx + c ) can be found using the formula ( x = -\frac{b}{2a} ). Once you determine the x-coordinate of the vertex, you can substitute it back into the function to find the corresponding y-coordinate. Therefore, the vertex is at the point ( \left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right) ). If the function is given as ( f(x) = x^2 + c ) (where ( a = 1 ) and ( b = 0 )), the vertex simplifies to ( (0, c) ).
In an isosceles triangle one base angle equals 73 find the size of the vertex angle?
jaj no se kompas jaj
Where could i find a vertex?
You can find a vertex wherever two lines (or line segments) meet.
