Practice Worksheet Graphing Quadratic Functions In Vertex Form ... - Free Printable

I'll solve this practice worksheet on graphing quadratic functions in vertex form. The worksheet contains six problems, each asking for the axis of symmetry, vertex, whether the parabola opens up or down, the slope to a point one unit from the vertex, and the y-intercept.

Let me work through each problem systematically.

Problem 1: $y = (x - 2)^2$

This is in vertex form $y = a(x - h)^2 + k$, where $(h, k)$ is the vertex.
- Vertex: $(2, 0)$
- Axis of Symmetry: $x = 2$
- Opens: Up (since $a = 1 > 0$)
- Slope to point one unit from vertex: For a parabola $y = a(x - h)^2 + k$, the slope from the vertex to a point one unit away horizontally is $2a$. Here, $a = 1$, so slope = $2(1) = 2$.
- Y-intercept: Set $x = 0$: $y = (0 - 2)^2 = 4$, so y-intercept is $(0, 4)$

Problem 2: $y = 3(x + 1)^2 + 5$

- Vertex: $(-1, 5)$
- Axis of Symmetry: $x = -1$
- Opens: Up (since $a = 3 > 0$)
- Slope to point one unit from vertex: $2a = 2(3) = 6$
- Y-intercept: Set $x = 0$: $y = 3(0 + 1)^2 + 5 = 3(1) + 5 = 8$, so y-intercept is $(0, 8)$

Problem 3: $y = 2(x + 1)^2 - 3$

- Vertex: $(-1, -3)$
- Axis of Symmetry: $x = -1$
- Opens: Up (since $a = 2 > 0$)
- Slope to point one unit from vertex: $2a = 2(2) = 4$
- Y-intercept: Set $x = 0$: $y = 2(0 + 1)^2 - 3 = 2(1) - 3 = -1$, so y-intercept is $(0, -1)$

Problem 4: $y = -2(x - 2)^2 - 3$

- Vertex: $(2, -3)$
- Axis of Symmetry: $x = 2$
- Opens: Down (since $a = -2 < 0$)
- Slope to point one unit from vertex: $2a = 2(-2) = -4$
- Y-intercept: Set $x = 0$: $y = -2(0 - 2)^2 - 3 = -2(4) - 3 = -8 - 3 = -11$, so y-intercept is $(0, -11)$

Problem 5: $y = \frac{1}{2}(x - 3)^2 + 2$

- Vertex: $(3, 2)$
- Axis of Symmetry: $x = 3$
- Opens: Up (since $a = \frac{1}{2} > 0$)
- Slope to point one unit from vertex: $2a = 2(\frac{1}{2}) = 1$
- Y-intercept: Set $x = 0$: $y = \frac{1}{2}(0 - 3)^2 + 2 = \frac{1}{2}(9) + 2 = 4.5 + 2 = 6.5$, so y-intercept is $(0, 6.5)$

Problem 6: $y = -(x + 2)^2 + 1$

- Vertex: $(-2, 1)$
- Axis of Symmetry: $x = -2$
- Opens: Down (since $a = -1 < 0$)
- Slope to point one unit from vertex: $2a = 2(-1) = -2$
- Y-intercept: Set $x = 0$: $y = -(0 + 2)^2 + 1 = -(4) + 1 = -3$, so y-intercept is $(0, -3)$

Now I'll fill in all the answers for the worksheet.

Completed Worksheet Answers:

1) $y = (x - 2)^2$
- Axis of Symmetry: $x = 2$
- Vertex: $(2, 0)$
- Opens: Up
- Slope to point one unit from the vertex: $2$
- Y-intercept: $(0, 4)$

2) $y = 3(x + 1)^2 + 5$
- Axis of Symmetry: $x = -1$
- Vertex: $(-1, 5)$
- Opens: Up
- Slope to point one unit from the vertex: $6$
- Y-intercept: $(0, 8)$

3) $y = 2(x + 1)^2 - 3$
- Axis of Symmetry: $x = -1$
- Vertex: $(-1, -3)$
- Opens: Up
- Slope to point one unit from the vertex: $4$
- Y-intercept: $(0, -1)$

4) $y = -2(x - 2)^2 - 3$
- Axis of Symmetry: $x = 2$
- Vertex: $(2, -3)$
- Opens: Down
- Slope to point one unit from the vertex: $-4$
- Y-intercept: $(0, -11)$

5) $y = \frac{1}{2}(x - 3)^2 + 2$
- Axis of Symmetry: $x = 3$
- Vertex: $(3, 2)$
- Opens: Up
- Slope to point one unit from the vertex: $1$
- Y-intercept: $(0, 6.5)$

6) $y = -(x + 2)^2 + 1$
- Axis of Symmetry: $x = -2$
- Vertex: $(-2, 1)$
- Opens: Down
- Slope to point one unit from the vertex: $-2$
- Y-intercept: $(0, -3)$