How To Find A Value In Vertex Form
HOW TO FIND THE MINIMUM OR MAXIMUM VALUE OF A FUNCTION IN VERTEX FORM
To observe the vertex course of the parabola, we use the concept completing the square method.
Vertex form of a quadratic function :
y = a(x - h)ii + k
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 upwardly, then it volition take minimum value
If a > 0, and then minimum value of f is f(h) = chiliad
Maximum value of parabola :
If the parabola is open downwards, then it will take maximum value.
If a < 0, then maximum value of f is f(h) = k
Practice Problems
Trouble 1 :
For the given functionf(x) = x 2 + 7x + 12
(a) Write f(x) in the form k(x + t)2 + r.
(b) Notice the value of ten where f(x) attains its minimum value or its maximum value.
(c) Detect the vertex of the graph of f.
Solution :
Let y =x ii + 7x + 12
y = x 2 + 2 ⋅ x ⋅(seven/2) + (7/ii)2 - (7/ii)2 + 12
y = (x + (vii/2))2 + 12
Past comparison it with vertex class, nosotros get the value of k . Since information technology is positive, the parabola is open upwards. So it volition minimum value.
(b) It has minimum value when ten = -7/2
(c) Vertex of the parabola is (-seven/2, 12)
Problem 2 :
For the given functionf(x) = 5x2 + 2x + one
(a) Write f(x) in the form k(x + t)2 + r.
(b) Notice the value of x where f(x) attains its minimum value or its maximum value.
(c) Discover the vertex of the graph of f.
Solution :
Let y = 5x2 + 2x + 1
y = 5(x2 + x) + 1
y = 5 [xtwo + 2 ⋅ x ⋅ (one/ii) + (1/two)2 - (one/two)ii] + 1
y = 5 [10 + (1/2)]ii - (1/4)] + 1
y = 5 [x + (1/ii)]2 - (5/4) + i
y = v [x + (one/2)]ii - (1/4)
By comparison it with vertex form, nosotros get the value of thousand . Since it is positive, the parabola is open upward. So information technology volition minimum value.
(b) It has minimum value when x = -one/2
(c) Vertex of the parabola is (-1/2, -1/iv)
Problem 3 :
For the given functionf(ten) = −2x2 + 5x − 2
(a) Write f(x) in the grade chiliad(ten + t)2 + r.
(b) Notice the value of x where f(x) attains its minimum value or its maximum value.
(c) Discover the vertex of the graph of f.
Solution :
Let y = −2xii + 5x − 2
y = −ii[10 ii - (five/2)10] − 2
y = −2[x2 - (5/two)10] − 2
y = -two [ten2 - 2 ⋅ ten ⋅ (v/4) + (5/4)ii - (5/4)2] - 2
y = -2 [x -(5/4)]2 - (25/sixteen)] - 2
y = -2 [x -(5/4)]2 + (25/8) - ii
y = -2 [x -(v/iv)]2 + (nine/8)
Past comparing it with vertex form, we get the value of thousand . Since it is negative, the parabola is open up downward. So it will maximum value.
(b) It has maximum value when ten = v/4
(c) Vertex of the parabola is (v/4, 9/eight)
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