00:07 Maxima /maximum Intro
01:03 Minima / minimum Intro
01:36 [Q1]: Find, using calculus, the maximum point of f(x), for f(x) = -x^2+x+6
02:52 [Q2]: Find the minimum point of the function f(x) = 3*x^2-6*x-8
Here we consider a standard form of the quadratic function as
f(x)=a*x^2+b*x+c,
where a, b and c are constants.
For value of 'a' being negative that is a less than 0,
the second derivative of the function is also negative as
2nd derivative = 2*a,
and the graph of the function is concave downward parabola . In this case the turning point of the parabola which is the
vertex of the parabola is a maximum point.
Further it is shown that the slope of the tangent line at the turning point is zero therefore the first derivative is zero at this point.
In order to find the maximum point we set first dertivative equal to zero and find the value of the x-coordinate.
Then we use this x-value in the given equation to find the corresponding y-value.
For the value of a being positive that is a greater than 0:
the second derivative of the quadratic function is also positive,
and the parabola is concave upward in this case.
Therefore the turning point which is the vertex of the parabola is a minimum point of the function.
Two exam questions are shown as examples:
01:36 [Q1]: Find, using calculus, the maximum point of f(x),
for f(x) = -x^2+x+6
02:52 [Q2]: Find the minimum point of the function f(x) = 3*x^2-6*x-8
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Leavingcert |#Quadratic #maxima #minima| thinkeccel
27 сен 2024
