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y= -5/49(x-9)^2+5

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14y ago

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Convert between standard and vertex form?

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What is the difference between standard form and vertex form?

The difference between standard form and vertex form is the standard form gives the coefficients(a,b,c) of the different powers of x. The vertex form gives the vertex 9hk) of the parabola as part of the equation.


How do you convert vertex form to quadratic form?

Do you have a specific vertex fraction? vertex = -b/2a wuadratic = ax^ + bx + c


How do you convert vertex form to standard form in algebra?

To convert a quadratic equation from vertex form, (y = a(x - h)^2 + k), to standard form, (y = ax^2 + bx + c), you need to expand the equation. Start by squaring the binomial: ( (x - h)^2 = x^2 - 2hx + h^2 ). Then, multiply by (a) and add (k) to obtain (y = ax^2 - 2ahx + (ah^2 + k)), where (b = -2ah) and (c = ah^2 + k). This results in the standard form of the quadratic equation.


How do you convert a vertex form equation of a parabola 2 standard form equation?

To convert a vertex form equation of a parabola, given as ( y = a(x - h)^2 + k ), to standard form ( y = ax^2 + bx + c ), expand the squared term: ( (x - h)^2 = x^2 - 2hx + h^2 ). Then, multiply through by ( a ) and combine like terms: ( y = ax^2 - 2ahx + (ah^2 + k) ). The coefficients ( a ), ( b = -2ah ), and ( c = ah^2 + k ) represent the standard form parameters.


How do you convert 144785 into standard form?

That already is in standard form.


What is the vertex of the parabola y equals -2x squared plus 12x -13?

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)


What are the benefits of writing a quadratic equation in vertex form and the benefits of writing a quadratic equation in standard form Name specific methods you can use while working with one form or?

Writing a quadratic equation in vertex form, ( y = a(x-h)^2 + k ), highlights the vertex of the parabola, making it easier to graph and identify key features like the maximum or minimum value. In contrast, standard form, ( y = ax^2 + bx + c ), is useful for quickly determining the y-intercept and applying the quadratic formula for finding roots. When working with vertex form, methods like completing the square can be employed to convert from standard form, while factoring or using the quadratic formula can be more straightforward when in standard form. Each form serves specific purposes depending on the analysis needed.


The vertex form of the equation of a parabola is . What is the standard form of the equation?

To convert the vertex form of a parabola, which is typically expressed as (y = a(x-h)^2 + k), into standard form (y = ax^2 + bx + c), you need to expand the equation. Start by squaring the binomial ((x-h)), which gives (x^2 - 2hx + h^2). Then, distribute the coefficient (a) and combine like terms to achieve the standard form. The resulting equation will be (y = ax^2 - 2ahx + (ah^2 + k)).


What different information do you get from vertex form and quadratic equation in standard form?

The graph of a quadratic function is always a parabola. If you put the equation (or function) into vertex form, you can read off the coordinates of the vertex, and you know the shape and orientation (up/down) of the parabola.


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.


What is the standard form of the equation of the parabola with vertex 00 and directrix y4?

Assuming the vertex is 0,0 and the directrix is y=4 x^2=0