Practice Solutions

Use substitution to write the quadratic equation in vertex form.  

1. $a=6, h=14, k=–8$
    $y=6{\left(x–14\right)}^2–8$

2. $a=–4, h=–9, k=–2$
    $y=–4{\left(x+9\right)}^2–2$

3. $a=\frac{1}{3}, h=7, k=10$
    $y=\frac{1}{3}{\left(x–7\right)}^2+10$

4. $a=8, h=0, k=5$
    $$\begin{gathered} y=8{\left(x–0\right)}^2+5 \\ y=8{\left(x\right)}^2+5 \end{gathered}$$

Identify a, h, and k. Name the vertex and the direction of the graph.

5. $y=2{\left(x–3\right)}^2+7$
   a = 2, h = 3, k = 7
   The vertex is (3, 7) and the graph opens upward because a is positive.

6. $y=\frac{1}{2}{\left(x+2\right)}^2+5$
   
    $a=\frac{1}{2}, h=–2, k=5$
   The vertex is (–2, 5) and the graph opens upward because a is positive.

Identify a, h, and k. Name the vertex and the direction of the graph.

7. $y={\left(x–6\right)}^2–8$
   a = 1, h = 6, k = –8
   The vertex is (6, –8) and the graph opens upward because a is positive.

8. $y=–{\left(x–4\right)}^2+10$
   a = –1, h = 4, k = 10
   The vertex is (4, 10) and the graph opens downward (reflected) because a is negative.

Describe the transformation from the parent function. Explain your reasoning.

9. $y={\left(x–2\right)}^2–4$
   The graph is open upward because a = 1, shifted right 2 spaces because h = 2, and shifted down 4 spaces because k = –4.

10. $y=–{\left(x+1\right)}^2+2$
    The graph is open downward because a = –1, shifted left 1 space because h = –1, and shifted up 2 spaces because k = 2.

11. $y={\left(x–3\right)}^2+12$
    The graph is open upward because a = 1, shifted right 3 spaces because h = 3, and shifted up 12 spaces because k = 12.

12. $y=–{\left(x+3\right)}^2–1$
    The graph is open downward because a = –1, shifted left 3 spaces because h = –3, and shifted down 1 space because k = –1.