tssm topic 1

tssm topic 1

curve sketching 1 - linear, quadratic, cubic and quartic functions

glossary

• factor theorem
• polynomial long division
• equation of linear equation using points
• angle of linear equation using slope

1. without using division, show that the first polynomial is divisible by the second

   1. $f(x)=x^3-x^2+x-1$ and $x=1$
      the factor theorem says that when we divide $f$ by $x-1$ the remainder is $f(1)$, which is equal to zero, which proves that it is divisible.
   2. show that $f(x)=2x^3-13x^2+27x-18$ and $2x-3$
      the factor theorem says that when we divide $f$ by $2x-3$ the remainder is $f(\frac{3}{2})$, which is equal to zero, which proves that it is divisible.

2. find the value of $k$ if the first polynomial is divisible by the second

   1. $f(x)=x^3-4x^2+x+k$ and $x-3$
      use the factor theorem
      since $x^3-4x^2+x+k$ is divisible by $x-3$, when we divide them the remainder is $f(3)$ which is zero
      since $f(3)=0$, use substitution to find $k=6$

3. $f(x)=x^3-(k+1)x^2-x+30$ and $x+3$
   use the factor theorem
   since $x^3-(k+1)x^2-x+30$ is divisible by $x+3$, when we divide them the remainder is $f(-3)$ which is zero
   since $f(-3)=0$, use substitution to find $k=-\frac{1}{3}$

4. factorize the following cubic functions

   1. factorize $f(x)=x^3-4x^2+x+6$
      use the factor theorem
      we can substitute values into $x$ and try to find a x-intercept
      through guessing, we find $f(3)=0$
      according to the factor theorem, $x-3$ is a factor of $f$
      to get the other factors we can divide $f$ by $x-3$ to get $x^2-x-2$
      $f(x)=(x-3)(x-2)(x+1)$

   2. factorize $f(x)=2x^3-3x^2-11x+6$
      use the factor theorem
      we can substitute values into $x$ and try to find a x-intercept
      through guessing, we find $f(-2)=0$
      according to the factor theorem, $x+2$ is a factor of $f$
      to get the other factors we can divide $f$ by $x+2$ to get $2x^2-7x+3$
      $f(x)=(x+2)(2x-1)(x-3)$

5. sketch the graphs of the following equations. state the domain and range.

   1. $3x+4y=12$
      domain is $\mathbb{R}$
      range is $\mathbb{R}$
      substitute $y=0$ to find x-intercept
      substitute $x=0$ to find x-intercept

   2. $3x-5y+12=0$
      domain is $\mathbb{R}$
      range is $\mathbb{R}$
      substitute $y=0$ to find x-intercept
      substitute $x=0$ to find x-intercept

6. find the equation of the line in the form of $ax+by+d$ in each of the following

   1. a line that has a slope of $\frac{1}{2}$ and passes through $(6,2)$
      we know that $y-2=\frac{1}{2}(x-6)$
      $y-2=\frac{x}{2}-3$
      $-\frac{x}{2}+y=-1$
   2. a line that passes through $(a,2)$ and $(4,-3)$
      we calculate that the slope is $\frac{5}{a-4}$
      we know that $y+3=\frac{5}{a-4}(x-4)$
      $y+3=\frac{5x-4}{a-4}$
      $5x-(a-4)y=8+3a$

7. a line joins the points $(-4,3)$ and $(-2,-1)$

   1. find the exact distance between the points
      $\sqrt{2^2+4^2}$
      $\sqrt{20}$
      $2\sqrt{5}$

   2. find the angle the line makes with the positive direction of the x-axis
      we find the slope
      $m=\frac{-4}{2}=-2$
      to find the acute angle the equation makes with the x-axis $\theta$
      $\theta =|{\tan }^{-1}(-2)|=63.43°$
      since the slope is negative, we subtract this from 180 to get answer
      $180-63.34=116.57°$

8. find all angles between the lines $y=2x-5$ and $y=-3x+5$ to 1 decimal place
   relative to the positive direction of the x-axis, the first equation has an angle of $|{\tan }^{-1}(2)|=63.43°$
   relative to the positive direction of the x-axis, the second equation has an angle of $180-|{\tan }^{-1}(-3)|=108.43°$
   the difference between the two is $45°$
   therefore, the angles between the lines are $45°$ and $135°$

9. find the solutions to the following equations

   1. $x^2+2x-15=0$
      $(x+5)(x-3)=0$
      $x=-5$ or $x=3$

   2. $8m^2=800$
      $m^2=100$
      $m=10$

   3. $x^2+10=7x$
      $(x-2)(x-5)=0$
      $x=2$ or $x=5$

   4. $7x^2=2(17x-12)$
      $7x^2-34x+24=0$
      $7x^2-6x-28x+24=0$
      $x(7x-6)+-4(7x-6)=0$
      $(x-4)(7x-6)=0$
      $x=4$ or $x=\frac{6}{7}$

10. sketch the graphs of the following stating domain and range

11. $f(x)=x^2+6x+8$
    domain is $\mathbb{R}$
    put into vertex form and complete the square
    $f(x)=x^2+6x+9-1$
    $f(x)=(x+3{)}^2-1$
    vertex is $(-3,-1)$
    range is $[-1,\infty )$
    find x-intercepts by setting $f(x)=0$
    $x^2+6x+8=0$
    $(x+2)(x+4)=0$
    $x=-2$ or $x=-4$
    find y-intercept by setting $x=0$
    $(0,8)$

12. $f(x)=(x-1{)}^2+4$
    domain is $\mathbb{R}$
    vertex is $(1,4)$
    range is $[4,\infty )$
    no x-intercepts
    find y-intercept by setting $x=0$
    $(0,5)$

13. $f(x)=-x^2-x+6$
    domain is $\mathbb{R}$
    put into vertex form and complete the square
    $f(x)=-x^2-x-0.25+6.25$
    $f(x)=-(x^2+x+0.25)+6.25$
    $f(x)=-(x+0.5{)}^2+6.25$
    vertex is $(-0.5,6.25)$
    range is $(-\infty ,6.25]$
    find x-intercepts by setting $f(x)=0$
    $-x^2-x+6=0$
    $x^2+x-6=0$
    $(x+3)(x-2)=0$
    $x=-3$ or $x=2$
    find y-intercept by setting $x=0$
    $(0,6)$

14. $f(x)=x^2-3x-2,-5\le x\le 3$
    domain is $[-5,3]$
    put into vertex form and complete the square
    $f(x)=x^2-3x+2.25-4.25$
    $f(x)=(x-1.5{)}^2-4.25$
    vertex is $(1.5,-4.25)$
    the highest value $f(x)$ can take is an $x$ value furthest away from 1.5, at $x=-5,f(x)=38$
    range is $[-4.25,38]$
    find x-intercepts by setting $f(x)=0$
    $x^2-3x-2=0$
    not able to be factored, use quadratic formula
    $x=\frac{3\pm \sqrt{9-4\cdot 1\cdot -2}}{4\cdot 1\cdot -2}$
    $x=\frac{3\pm \sqrt{17}}{2}$
    apply domain restrictions
    $x=\frac{3-\sqrt{17}}{2}$
    find y-intercept by setting $x=0$
    $(0,-2)$

15. let $f:\mathbb{R}\to \mathbb{R},f(x)=x^2+5x-8$

    1. write the function in turning point form, state coordinates of the turning point

    to find the turning point
    $f(x)=x^2+5x+(2.5{)}^2-(2.5{)}^2-8$
    $f(x)=(x+2.5{)}^2-14.25$
    so now we know that the turning point is at $(-2.5,-14.25)$

    1. use the discriminant to determine the number of x-intercepts

    the discriminant is $57$ which is greater than zero, meaning that there are 2 x-intercepts

    1. state the domain and range. sketch the function

    domain is $\mathbb{R}$
    range is $[-14.25,\infty )$

    find the x-intercepts using the quadratic formula
    $x=\frac{-(\sqrt{57}+5)}{2}$ or $x=\frac{\sqrt{57}-5}{2}$

    find y-intercept by setting $x=0$
    $(0,-8)$

    1. solve the equation $2x^3-7x^2+2x+3=0$

16. sketch the following graphs showing all relevant information

    1. $f(x)=(x-2{)}^3-2$
       find x-intercepts by setting $f(x)=0$
       find y-intercept by setting $x=0$

    2. $f(x)=(x+3)(1-x)(x-4)$
       find x-intercepts by setting $f(x)=0$
       find y-intercept by setting $x=0$

    3. $f(x)=(x+2{)}^2(x-3)$
       find x-intercepts by setting $f(x)=0$
       find y-intercept by setting $x=0$
       repeated factor is a turning point

    4. $f(x)=x^3-5x^2+x+10$
       we need to factor this using the factor theorem

17. sketch the following graphs showing all relevant information

18. $f(x)=(x+2)(x-2)(x+4)(x-5)$

    1. $f(x)=x^2(x+4)(5-x)$
       repeated factor is a turning point

    2. $f(x)=(x+2{)}^2(x-6{)}^2$
       repeated factors are a turning points

    3. $f(x)=(x+2{)}^3(x-6)$
       cubed factor is an inflection point

19. for the simultaneous equations $(m-3)x+8y=10$ and $5x+(m+3)y=11$, find the values of $m$ for which the equations have:

    1. no solutions
    2. unique solutions
    3. infinite solutions