Answer to Question #274705 in Algebra for Bjoy

Question

Sketch the graph of the functions:

1. f(x) = 5

2. h(x) = 3x + 2

3. q(x) = x2 + 6x -7

4. k(x) =

5. h ₒ g = 2x3 + x2 – 10x - 5

6. g ₒ f = (2x + 1)(x – 3) = 2x2 – 5x -3

Answer 1

1. f(x)=5

This function has the constant value equal to 5, the graph is a horizontal line at this height

1. f(x)=3$\cdot x+2$

This is the graph of a straight line. For drawing it is enough any two points on it, for example, $x_1=0,y_1=3\cdot x_1+2=3\cdot 0+2=2,\space P_1(0,2)$

$x_2=1,y_2=3\cdot x_2+2=3\cdot 1+2=5,\space P_2(1,5)$

1. q(x)=$x^2+6\cdot x-7$

This is a graph of a quadratic trinomial or parabola. It is useful to find its roots by solving an equation $x^2+6\cdot x-7=0$ . We have

$x_1=\frac{-6-\sqrt{6^2-4\cdot 1\cdot (-7)}}{2\cdot 1}=\frac{-6-\sqrt{64}}{2\cdot 1}=\frac{-6-8}{2}=-7$

$x_2=\frac{-6+\sqrt{6^2-4\cdot 1\cdot (-7)}}{2\cdot 1}=\frac{-6+\sqrt{64}}{2\cdot 1}=\frac{-6-+8}{2}=1$

Thus we have two points $P_1(-7,0),P_2(1,0)$ on the graph. The middle point is$x_3=\frac{-7+1}{2}=-3,y_3=(-3)^2+6\cdot(-3)-7=-16$

Using these 3 sample points we may sketch the graph:

4 $k(x)=x^3$

This is a cubic parabola. For drawing its graph we create the table of sample values such as

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} x & -2 & -1 & 0 &1 &2\\ \hline x^3 & -8 & -1 &0 &1&8\\ \hline \end{array}$

and draw a curve through them

5. h ₒ g =P(x)= 2x³ + x² – 10x - 5

This line of the polynomial of 3-th degree

First derivative dP(x)/dx=8$\cdot x^2+2x-10=0 ~\harr x_1=1,x_2=\frac{5}{4}$

Second derivative $\frac{d^2P(x)}{dx^2}=16\cdot x+2$

$\frac{d^2P(x)}{dx^2}(1)=16\cdot 1+2=18>0\rarr x_1=1$ is a point of local minimum. P(1)=-12;

$\frac{d^2P(x)}{dx^2}(-\frac{5}{4})=-20+2=-18>0\rarr x_2=-\frac{5}{4}$ is the point of local maximum.

$P(-\frac{5}{4})=-\frac{-125}{32}+\frac{25}{16}+\frac{50}{4}-5=\frac{165}{32}$

On $(-\infty ,-\frac{5}{4})$ P(x) increases, on($-\frac{5}{4},1)$ deceases and on $(1.\infty )$ increases again.

P(-2.5)=-5,P(2.5)=7.5.P(0)=-5

Using analyzed information we sketch the graph:

6. g ₒ f =Q(x)= (2x + 1)(x – 3) = 2x² – 5x -3

This is a parabola. Firstly we find its roots

$x_1=\frac{5-\sqrt{49}}{4}=-0.5, x_2=\frac{5+\sqrt{49}}{4}=3,$

the vertex of the parabola is in the middle point $x_3=\frac{-0.5+3}{2}=1.25,y_3=2\cdot 1.25^2-5\cdot 1.25-3$ =-6.125

Q(-1)=5,Q(4)=9. Using this information we draw the graph: