Find the minimum value of the function

Mathematics

1 answer:

Thepotemich [5.8K]3 years ago

7 0

$\bf \textit{ vertex of a vertical parabola, using coefficients}\\\\
\begin{array}{lccclll}
g(x) = &{{ 2}}x^2&{{ +4}}x&{{ -5}}\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad 
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)
\\\\\\
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad \stackrel{minimum}{-5-\cfrac{4^2}{4(2)}}\right)$

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Given the three vertices W(−1, 4), X(−3, −2), and Y(3, −4), what are the coordinates of Z that make quadrilateral WXYZ a square?

prohojiy [21]

Answer:

I don't see how the three existing points could ever become a square with the addition of a foiurth point.

Step-by-step explanation:

See the attached image.

A square would require that all angles be 90 degrees. The given points are the top three points on the graph. If we enter the two equations that intersect these points (blue and black lines), we can see that the angle on top is not 90 degrees. I can't see that this could ever be a square with a fourth point, z. I did find a value for z that make the four points a parallelogram.