HOW TO SKETCH SQUARE ROOT FUNCTION

Sketch the graphs of the following functions.

Problem 1 :

y  =  x√(4-x)

Solution :

Domain and range :

Possible values of x are (-∞, 3) ==>  Domain

y values are (-∞, 4) ==>  Range

Intercepts :

To find x intercept, put y  =  0

  x√(4-x)  =  0

x intercepts are 0 and 4. 

y-intercept :

Put x  =  0

y  =  0

Critical points :

f(x)  =  x√(4-x)

f'(x)  =  x(1/2√(4-x))(-1) + √(4-x) (1)

f'(x)  =  -x/2√(4-x) + √(4-x)

f'(x)  =  [-x+2(4-x)]/2√(4-x)

f'(x)  =  (-x+8-2x)/2√(4-x)

f'(x)  =  (8-3x)/2√(4-x)

f'(x)  =  0

(8-3x)/2√(4-x)  =  0

3x  =  8

x  =  8/3

Plotting these points on the number line and dividing into intervals, we get (-∞, 8/3) and (8/3, ∞).

Local extrema :

f'(x)  =  (8-3x)/2√(4-x)

u  =  8-3x and v  =  2√(4-x)

u'  =  -3 and v'  =  -1/√(4-x)

f''(x)  =  [2√(4-x)(-3)-(8-3x)(-1/√(4-x))] /4(4-x)

f''(x)  =  (3x-16)/4(4-x)√(4-x)

f''(8/3)  =  - < 0 (Maximum)

Maximum value at x  =  8/3 :

y  =  x√(4-x)

y  =  (8/3)√(4-(8/3))

y  =  (8/3)√(4/3)

y  =  16/3√3

So, the maximum point is (8/3, 16/3√3).

Intervals of concavity :

f''(x)  =  (3x-16)/4(4-x)√(4-x)

f''(x)  =  0

3x-16  =  0

x  =  16/3

It is not in the domain, so it has no real value on.

Point of inflection :

There is no point of inflection.

Asymptotes :

As x ->∞, y -> ±∞

Identify the domain and range of each. Then sketch the graph.

Problem 2 :

y = -2√(x + 1)

Solution :

Domain and range :

Possible values of x are (-1, ∞) ==>  Domain

y values are (0, -∞) ==>  Range

Intercepts :

To find x intercept, put y  =  0

-2√(x + 1) =  0

√(x + 1) =  0

x + 1 = 0

x = -1

x intercept is at x = -1.

y-intercept :

Put x  =  0

y = -2√(x + 1)

y = -2√(0 + 1)

y = -2

Comparing with y = a√(x - h) + k,

we get move the graph of parent function 1 unit to the left and there is no vertical translation.

Reflection about x-axis and 2 units of vertical stretch.

By applying some inputs, we will get the outputs.

y = -2√(x + 1)

The set of ordered pairs are (-1, 0) (0, -2) (1, -4) (2, -2√3) and (3, -4).

How would each of the following graphs change in relation to the parent graph?

Problem 3 :

y = √(x - 3)

Solution :

Domain and range :

(x - 3) ≥ 0

x  ≥ 3

Possible values of x are (3, ∞) ==>  Domain

y values are (0, ∞) ==>  Range

Intercepts :

To find x intercept, put y  =  0

√(x - 3) =  0

x - 3 = 0

x = 3

x intercept is at x = 3.

y-intercept :

Put x  =  0

y = √(0 - 3)

There is no x-intercept

Comparing with y = a√(x - h) + k,

we get move the graph of parent function should be moved 3 units to the right.

y = √(x - 3)

Problem 4 :

y = 2√(x + 2) - 3

Solution :

Domain and range :

(x + 2) ≥ 0

x ≥ 2

Possible values of x are (2, ∞) ==>  Domain

y values are (0, ∞) ==>  Range

Intercepts :

To find x intercept, put y  =  0

2√(x + 2) - 3 =  0

2√(x + 2) = 3

√(x + 2) = 3/2

x + 2 = 9/4

x = (9/4) - 2

x = 1/4

x intercept is at x = 1/4

y-intercept :

Put x  =  0

y = 2√(0 + 2) - 3

y = 2√2 - 3

= 2(1.414) - 3

= 2.828 - 3

= -0.172

y-intercept is (0, -0.172)

Comparing with y = a√(x - h) + k,

we get move the graph of parent function,

• Vertical stretch of 2 units
• Moving the graph 2 units left
• Moving the graph 3 units down.

y = 2√(x + 2) - 3

The ordered pairs are (-2, -3) (-1, -1) (0, 2√2 - 3) (6, 4√2 - 3) (7, 3).

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