## FINDING MAXIMUM AND MINIMUM VALUE OF QUADRATIC FUNCTION

Finding Maximum and Minimum Value of Quadratic Function :

Here we are going to see, how to find the maximum and minimum value of quadratic function.

## Method to Find Maximum and Minimum Value of Quadratic Function

The vertex of a parabola is the point where the line of symmetry of the parabola intersects the parabola.

Let f be a quadratic function with standard form

y = a(x - h)2 + k

the maximum and minimum value of f occurs at x = h

In order to find the maximum or minimum value of quadratic function, we have to convert the given quadratic equation in the above form.

Minimum value of parabola :

If the parabola is open upward, then it will have minimum value If a > 0, then minimum value of f is f(h)  =  k

Maximum value of parabola :

If the parabola is open downward, then it will have maximum value. If a < 0, then maximum value of f is f(h)  =  k

## Finding Maximum and Minimum Value of Quadratic Function

Question 1 :

Suppose f , g, and h are defined by

f(x) = −x2, g(x) = −2x2, h(x) = −2x2 + 1.

(a) Sketch the graphs of f , g, and h on the interval [−1, 1].

(b) Find the vertex of the graph of f , the graph of g, and the graph of h.

(d) What is the maximum value of f ? The maximum value of g? The maximum value of h?

Solution :

Graph of f(x) = −x2 : Graph of f(x) = −2x: Graph of f(x) = −2x2 + 1 (b)  The vertex of the parabola f(x) = −x2 is at (0, 0). Hence the maximum value is 0.

The vertex of the parabola f(x) = −2x2 is at (0, 0). Hence the maximum value is 0.

The vertex of the parabola f(x) = −2x2 + 1 is at (0, 1). Hence the maximum value is 1.

Question 2 :

The function defined by

f(x) = x2 + 6x + 11.

(a) For what value of x does f(x) attain its minimum value?

(b) Sketch the graph.

(c)  Find the vertex of the graph of f .

Solution :

In order to convert the given quadratic function into vertex form, let us use the method completing the square method.

Let y = x2 + 6x + 11

y - 11  =  x2 + 6x

y - 11  =  x2 + 2 ⋅ ⋅ 3 + 32 - 32

y - 11  =  (x + 3)2 - 9

y  =  (x + 3)2 - 9 + 11

y  =  (x + 3)2 + 2

y = a(x - h)2 + k

a =  1 > 0, the parabola is open upward, then it will have minimum value. Vertex of parabola (h, k) is (-3, 2)

When x = -3, f(x) attains its minimum value.

(b)  After having gone through the stuff given above, we hope that the students would have understood "Finding Maximum and Minimum Value of Quadratic Function".

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