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
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) = −2x2 :
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 ⋅ x ⋅ 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)
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