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NotSoSmart  Nov 16, 2017

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Standard form of a quadratic is in the form \(ax^2+bx+c\), and vertex form is in the form of \(a(x-h)^2+k\).

 

a) The method of factoring requires a quadratic to be in standard form in order to use its procedure, so standard form is desired.

 

b) Vertex form is friendlier when it comes to graphing. One can quickly identify the vertex (hence its name vertex form), graph the points to the right, and then reflect about the axis of symmetry. Computationally speaking, it seems to be easier to indentify coordinates with vertex form. 

 

c) As aforementioned, vertex form allows one to identify the vertex simply by looking at the equation because \((h,k)\) is the vertex. In standard form, one would have to calculate \(\left(-\frac{b}{2a},f\left(-\frac{b}{2a}\right)\right)\), which requires more work.

 

d) In order to solve a quadratic using its corresponding formula, the formula requires the function be written in standard form. 

TheXSquaredFactor  Nov 17, 2017
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 #1
avatar+1493 
+2
Best Answer

Standard form of a quadratic is in the form \(ax^2+bx+c\), and vertex form is in the form of \(a(x-h)^2+k\).

 

a) The method of factoring requires a quadratic to be in standard form in order to use its procedure, so standard form is desired.

 

b) Vertex form is friendlier when it comes to graphing. One can quickly identify the vertex (hence its name vertex form), graph the points to the right, and then reflect about the axis of symmetry. Computationally speaking, it seems to be easier to indentify coordinates with vertex form. 

 

c) As aforementioned, vertex form allows one to identify the vertex simply by looking at the equation because \((h,k)\) is the vertex. In standard form, one would have to calculate \(\left(-\frac{b}{2a},f\left(-\frac{b}{2a}\right)\right)\), which requires more work.

 

d) In order to solve a quadratic using its corresponding formula, the formula requires the function be written in standard form. 

TheXSquaredFactor  Nov 17, 2017

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