SKETCHING THE GIVEN CURVE PRACTICE PROBLEMS

When we are sketching the graph of functions either by hand or through any graphing software we cannot show the entire graph. Only a part of the graph can be sketched.

A crucial question is which part of the curve we need to show and how to decide that part. To decide on this we use the derivatives of functions. We enlist few guidelines for determining a good viewing rectangle for the graph of a function. They are :

(i) The domain and the range of the function.

(ii) The intercepts of the cure (if any).

(iii) Critical points of the function.

(iv) Local extrema of the function.

(v) Intervals of concavity.

(vi) Points of inflections (if any).

(vii) Asymptotes of the curve (if exists)

Answer Key

1)

Since it is a polynomial function, all real values are domain and range.

1 and -2 are x-intercepts.

y-intercept is y  =  -2/3

Critical points are x  =  -1, 1

Decreasing on (-∞, -1) and (1, ∞).

Increasing on (-1, 1)

local extrema :

f''(-1)  =  2 > 0 (Minimum)

f''(1)  =  -2 < 0 (Maximum)

Concavity :

At (-∞, 0), concave up

At (0, ∞), concave down

Point of inflection :

point of inflection is (0, -2/3).

It will have only slant asymptote.

2)  Critical numbers are 0 and 2.

Decreasing on (-∞, 0)

increasing on (0, 2)

Decreasing on (2, ∞)

Concave up on (-∞, 1), Concave down on (1, ∞)

Point of inflection at x = 1

maximum at (2, 4/3).

3)  At x < -2 increasing

at -2 < x < 3 decreasing

At x > 3 increasing.

4) At x < -2 decreasing

At -2 < x < 0 increasing

At x > 0 decreasing

Answer Key

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

y values are (-∞, 4) ==>  Range

x intercepts are 0 and 4.

y-intercept is at y  =  0

At (-∞, 8/3) increasing

At (8/3, ∞) decreasing.

the maximum point is (8/3, 16/3√3).

There is no point of inflection.

As x ->∞, y -> ±∞

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

y values are (0, -∞) ==>  Range

Intercepts :

x intercept is at x = -1.

y-intercept :

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)

3)  Domain and range :

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

y values are (0, ∞) ==>  Range

Intercepts :

x intercept is at x = 3.

y-intercept :

There is no y-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)

4)  

Domain and range :

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

y values are (0, ∞) ==>  Range

Intercepts :

x intercept is at x = 1/4

y-intercept :

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.

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