What are the coordinates of the vertex for the function shown below ?
1 answer:
Answer:
Step-by-step explanation:
We can answer this in two ways: Differentiation and Graphing.
<u>Differentitate:</u>
The first derivative of a function yields a function that provides the slope for any point on the line of the original function. The slope of a vertex is zero, so we can set the first derivative to 0 and solve for x.
f(x) = 3(x-1)^2 + 4
f(x) = 3(x-1)(x-1) + 4
f(x) = 3(x^2 - 2x + 1) + 4
f(x) = 3x^2 - 6x + 7
f'(x) = 6x -6
Set this = 0 and find x:
0 = 6x -6
x = 1
The value of y when x=1 in the original equation is:
f(1) = 3(1-1)^2 + 4
y = 4
The vertex is (1,4)
<u>Graph:</u>
You can use DESMOS to plot the function. The result is attached. Look for the vertex and read the coordinates. (1,4) seems to work.
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Answer:
1.2 cm
Step-by-step explanation:
Quadrilateral circumscribing a circle is a quadrangle whose sides are tangent to a circle inside it (see attached diagram).
The area of circumscribed quadrilateral is
where is semi-perimeter and r is radius of inscribed circle.
In your case,
If a quadrilateral is circumscribed over the circle, then the sum of opposite sides is equal, so
so
Now
