# Exercises - Parabolas

1. Write each quadratic function in the form $y=a(x-h)^2+k$ and sketch its graph.

1. $y=x^2-3x$   $\ans{ \begin{array}{rcl} y &=& x^2-3x\\ &=& (x^2 - 3x + \frac{9}{4}) - \frac{9}{4}\\ &=& (x - \frac{3}{2})^2 - \frac{9}{4} \end{array}}$

2. $y=2x^2-12x+22$   $\ans{ \begin{array}{rcl} y &=& 2x^2-12x+22\\ &=& 2 (x^2 - 6x + 9) - 18 + 22\\ &=& 2 (x - 3)^2 + 4 \end{array}}$

3. $y=-\frac{1}{2} x^2 + x + \frac{5}{2}$   $\ans{ \begin{array}{rcl} y &=& -\frac{1}{2} x^2 + x + \frac{5}{2}\\ &=& -\frac{1}{2} ( x^2 - 2x + 1 ) + \frac{1}{2} + \frac{5}{2}\\ &=& -\frac{1}{2} (x-1)^2 + 3 \end{array}}$

4. $y = x^2 + 3x + \frac{5}{2}$   $\ans{\begin{array}{rcl} y &=& x^2 + 3x + \frac{5}{2}\\ &=& (x^2 + 3x + \frac{9}{4}) - \frac{9}{4} + \frac{5}{2}\\ &=& (x+ \frac{3}{2})^2 + \frac{1}{4} \end{array}}$

5. $y=-2x^2+3x-1$   $\ans{\begin{array}{rcl} y &=& -2x^2+3x-1\\ &=& -2(x^2-\frac{3}{2} x + \frac{9}{16}) + \frac{9}{8} - 1\\ &=& -2(x-\frac{3}{4})^2 + \frac{1}{8} \end{array}}$

6. $y=-3x^2+2x$   $\ans{\begin{array}{rcl} y &=& -3x^2+2x\\ &=& -3(x^2-\frac{2}{3} x + \frac{1}{9}) + \frac{1}{3}\\ &=& -3(x-\frac{1}{3})^2 + \frac{1}{3} \end{array}}$

2. Identify the vertex, axis of symmetry, $y$-intercept, $x$-intercepts, and direction of opening (up or down) of each parabola, then sketch the graph.

1. $y=x^2-3$   $\ans{ \begin{array}{l} \textrm{vertex: } (0,-3);\\ y\textrm{-intercept: } (0,-3);\\ \textrm{Solve } 0 = x^2 - 3 \textrm{ to find}\\ x\textrm{-intercepts: } (\pm \sqrt{3},0);\\ \textrm{opens up}\\ \end{array}}$

2. $f(x)=x^2- \pi x$   $\ans{ \begin{array}{l} \begin{array}{rcl} y &=& x^2 - \pi x\\ &=& (x^2 - \pi x + \frac{\pi^2}{4}) - \frac{\pi^2}{4}\\ &=& (x - \frac{\pi}{2})^2 - \frac{\pi^2}{4} \end{array}\\\\ \begin{array}{l} \textrm{vertex: } (\frac{\pi}{2}, -\frac{\pi^2}{4});\\ y\textrm{-intercept: } (0,0);\\ \textrm{Note, } y = x(x-\pi). \textrm{ Solving this}\\ \textrm{where } y=0 \textrm{ we find}\\ x\textrm{-intercepts: } (0,0), (\pi, 0);\\ \textrm{opens up} \end{array} \end{array}}$

3. $f(x)=x^2+6x+9$   $\ans{ \begin{array}{l} f(x) = (x+3)^2\\\\ \textrm{vertex: } (-3,0);\\ y\textrm{-intercept: } (0,9);\\ x\textrm{-intercept: }(-3,0);\\ \textrm{opens up}\\ \end{array}}$

4. $f(x)=(x-3)^2-4$   $\ans{ \begin{array}{l} \textrm{vertex: } (3,4);\\ y\textrm{-intercept: } (0,5);\\ \textrm{To find }x\textrm{-intercepts, we solve for where }f(x) = 0:\\\\ \begin{array}{rcl} (x-3)^2 - 4 &=& 0\\ x^2 - 6x + 9 - 4 &=& 0\\ x^2 - 6x + 5 &=& 0\\ (x-5)(x-1) &=& 0\\ x &=& 5 \textrm{ or } 1\\ \end{array}\\\\ x\textrm{-intercepts: } (5,0),(1,0)\\ \textrm{opens up}\\ \end{array}}$

5. $y=-3(x-2)^2+12$   $\ans{ \begin{array}{l} \textrm{vertex: } (2,12);\\ y\textrm{-intercept: } (0,0)\\ \textrm{To find }x\textrm{-intercepts, we solve for where }f(x) = 0:\\\\ \begin{array}{rcl} -3(x-2)^2+12 &=& 0\\ -3(x^2 - 4x + 4) + 12 &=& 0\\ -3x^2 + 12x &=& 0\\ -3x(x-4) &=& 0\\ x &=& 0 \textrm{ or } 4\\ \end{array}\\\\ x\textrm{-intercepts: } (0,0), (4,0)\\ \textrm{opens down}\\ \end{array}}$

6. $y=-2x^2+4x+1$   $\ans{ \begin{array}{l} \begin{array}{rcl} y &=& -2x^2+4x+1\\ &=& -2(x^2-2x+1) + 2 + 1\\ &=& -2(x-1)^2+3\\ \end{array}\\\\ \textrm{vertex: } (1,3)\\ y\textrm{-intercept: } (0,1)\\ \textrm{To find }x\textrm{-intercepts, we solve for where }f(x) = 0:\\\\ \begin{array}{rcl} -2(x-1)^2+3&=&0\\ (x-1)^2 &=& \frac{3}{2}\\ x - 1 &=& \pm \sqrt{\frac{3}{2}}\\ x &=& 1 \pm \sqrt{\frac{3}{2}}\\ x &=& \displaystyle{\frac{2 \pm \sqrt{6}}{2}} \end{array}\\\\ x\textrm{-intercepts: } \displaystyle{\left( \frac{2 \pm \sqrt{6}}{2}, 0 \right)}\\ \textrm{opens down}\\ \end{array}}$