- - What is the vertex of the quadratic function f(x) = (x - 6)(x + 2)?
2 answers:
The Standard Form: when a, b and c are real numbers and a must not be 0 (a ≠ 0)
We can convert it in the vertex form which is
Right now, the function takes form of intercepts, we have to convert it to the standard form as we will convert the standard form to vertex.
Distributed
At this part, we can complete the square inside as we'll get (x-h)^2 and k
The vertex is at (2, -16) [The vertex is at (-h, k) so (-(-2), -16) as we get (2, -16)]
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We first obtain the equation of the lines bounding R.
For the line with points (0, 0) and (8, 1), the equation is given by:
For the line with points (0, 0) and (1, 8), the equation is given by:
For the line with points (8, 1) and (1, 8), the equation is given by:
The Jacobian determinant is given by
The integrand x - 3y is transformed as 8u + v - 3(u + 8v) = 8u + v - 3u - 24v = 5u - 23v
Therefore, the integration is given by:
![63 \int\limits^1_0 \int\limits^{1}_0 {(5u-23v)} \, dudv =63 \int\limits^1_0\left[\frac{5}{2}u^2-23uv\right]^{1}_0 \\ \\ =63\int\limits^1_0(\frac{5}{2}-23v)dv=63\left[\frac{5}{2}v-\frac{23}{2}v^2\right]^1_0=63\left(\frac{5}{2}-\frac{23}{2}\right) \\ \\ =63(-9)=|-576|=576](https://tex.z-dn.net/?f=63%20%5Cint%5Climits%5E1_0%20%5Cint%5Climits%5E%7B1%7D_0%20%7B%285u-23v%29%7D%20%5C%2C%20dudv%20%3D63%20%5Cint%5Climits%5E1_0%5Cleft%5B%5Cfrac%7B5%7D%7B2%7Du%5E2-23uv%5Cright%5D%5E%7B1%7D_0%20%5C%5C%20%20%5C%5C%20%3D63%5Cint%5Climits%5E1_0%28%5Cfrac%7B5%7D%7B2%7D-23v%29dv%3D63%5Cleft%5B%5Cfrac%7B5%7D%7B2%7Dv-%5Cfrac%7B23%7D%7B2%7Dv%5E2%5Cright%5D%5E1_0%3D63%5Cleft%28%5Cfrac%7B5%7D%7B2%7D-%5Cfrac%7B23%7D%7B2%7D%5Cright%29%20%5C%5C%20%20%5C%5C%20%3D63%28-9%29%3D%7C-576%7C%3D576)
For the line with points (0, 0) and (8, 1), the equation is given by:
For the line with points (0, 0) and (1, 8), the equation is given by:
For the line with points (8, 1) and (1, 8), the equation is given by:
The Jacobian determinant is given by
The integrand x - 3y is transformed as 8u + v - 3(u + 8v) = 8u + v - 3u - 24v = 5u - 23v
Therefore, the integration is given by: