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Step-by-step solution for: How To Convert Standard Form To Vertex Form - Fill Online ...
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Step-by-step solution for: How To Convert Standard Form To Vertex Form - Fill Online ...
Since the user has not selected any of the optional features (ExplainAnswer, TeachTopic, GeneratePractice), I will proceed directly to solving the problems on the worksheet and providing the final answers.
The task is to convert equations of parabolas into Vertex Form.
The standard vertex form of a parabola is:
$$y = a(x - h)^2 + k$$
where $(h, k)$ is the vertex.
I will solve each of the 14 problems step-by-step.
Problem 1: $y = x^2 + 16x + 71$
1. Identify $a=1$, $b=16$, $c=71$.
2. Find the x-coordinate of the vertex: $h = -b / 2a = -16 / 2 = -8$.
3. Find the y-coordinate by plugging $h$ back in: $k = (-8)^2 + 16(-8) + 71 = 64 - 128 + 71 = 7$.
4. Vertex is $(-8, 7)$.
5. Equation: $y = (x - (-8))^2 + 7 \rightarrow y = (x + 8)^2 + 7$.
Problem 2: $y = x^2 - 2x - 5$
1. $a=1, b=-2, c=-5$.
2. $h = -(-2) / 2 = 1$.
3. $k = (1)^2 - 2(1) - 5 = 1 - 2 - 5 = -6$.
4. Vertex is $(1, -6)$.
5. Equation: $y = (x - 1)^2 - 6$.
Problem 3: $y = x^2 - 14x + 59$
1. $a=1, b=-14, c=59$.
2. $h = -(-14) / 2 = 7$.
3. $k = (7)^2 - 14(7) + 59 = 49 - 98 + 59 = 10$.
4. Vertex is $(7, 10)$.
5. Equation: $y = (x - 7)^2 + 10$.
Problem 4: $y = x^2 + 9x + 170$
1. $a=1, b=9, c=170$.
2. $h = -9 / 2 = -4.5$.
3. $k = (-4.5)^2 + 9(-4.5) + 170 = 20.25 - 40.5 + 170 = 149.75$.
4. Vertex is $(-4.5, 149.75)$.
5. Equation: $y = (x + 4.5)^2 + 149.75$.
Problem 5: $y = x^2 - 12x + 46$
1. $a=1, b=-12, c=46$.
2. $h = -(-12) / 2 = 6$.
3. $k = (6)^2 - 12(6) + 46 = 36 - 72 + 46 = 10$.
4. Vertex is $(6, 10)$.
5. Equation: $y = (x - 6)^2 + 10$.
Problem 6: $y = x^2 + 4x$
1. $a=1, b=4, c=0$.
2. $h = -4 / 2 = -2$.
3. $k = (-2)^2 + 4(-2) = 4 - 8 = -4$.
4. Vertex is $(-2, -4)$.
5. Equation: $y = (x + 2)^2 - 4$.
Problem 7: $y = x^2 - 6x + 5$
1. $a=1, b=-6, c=5$.
2. $h = -(-6) / 2 = 3$.
3. $k = (3)^2 - 6(3) + 5 = 9 - 18 + 5 = -4$.
4. Vertex is $(3, -4)$.
5. Equation: $y = (x - 3)^2 - 4$.
Problem 8: $y = (x + 5)(x + 4)$
1. First expand to standard form: $y = x^2 + 4x + 5x + 20 \rightarrow y = x^2 + 9x + 20$.
2. $a=1, b=9, c=20$.
3. $h = -9 / 2 = -4.5$.
4. $k = (-4.5)^2 + 9(-4.5) + 20 = 20.25 - 40.5 + 20 = -0.25$.
5. Vertex is $(-4.5, -0.25)$.
6. Equation: $y = (x + 4.5)^2 - 0.25$.
Problem 9: $\frac{1}{2}(x + 4) = (y - 2)^2$
1. This is a horizontal parabola. We need to isolate $y$.
2. Multiply by 2: $x + 4 = 2(y - 2)^2$.
3. Divide by 2: $\frac{1}{2}(x + 4) = (y - 2)^2$.
4. Take square root? No, usually vertex form for horizontal is $x = a(y-k)^2 + h$.
5. Let's rearrange to $x = \dots$
6. $x + 4 = 2(y-2)^2 \rightarrow x = 2(y-2)^2 - 4$.
7. Vertex is $(-4, 2)$.
8. Equation: $x = 2(y - 2)^2 - 4$.
Problem 10: $6x^2 + 12x + y + 13 = 0$
1. Isolate $y$: $y = -6x^2 - 12x - 13$.
2. $a=-6, b=-12, c=-13$.
3. $h = -(-12) / (2 \cdot -6) = 12 / -12 = -1$.
4. $k = -6(-1)^2 - 12(-1) - 13 = -6 + 12 - 13 = -7$.
5. Vertex is $(-1, -7)$.
6. Equation: $y = -6(x + 1)^2 - 7$.
Problem 11: $162x + 72y = -9y^2$
1. Rearrange to solve for $x$ (since $y^2$ term exists).
2. $162x = -9y^2 - 72y$.
3. Divide by 162: $x = \frac{-9}{162}y^2 - \frac{72}{162}y$.
4. Simplify fractions: $x = -\frac{1}{18}y^2 - \frac{4}{9}y$.
5. Find vertex y-coordinate: $k = -b / 2a = -(-4/9) / (2 \cdot -1/18) = (4/9) / (-1/9) = -4$.
6. Find vertex x-coordinate: Plug $y=-4$ into equation.
$x = -\frac{1}{18}(-4)^2 - \frac{4}{9}(-4) = -\frac{16}{18} + \frac{16}{9} = -\frac{8}{9} + \frac{16}{9} = \frac{8}{9}$.
7. Vertex is $(8/9, -4)$.
8. Equation: $x = -\frac{1}{18}(y + 4)^2 + \frac{8}{9}$.
Problem 12: $x^2 - 12x + y + 40 = 0$
1. Isolate $y$: $y = -x^2 + 12x - 40$.
2. $a=-1, b=12, c=-40$.
3. $h = -12 / (2 \cdot -1) = 6$.
4. $k = -(6)^2 + 12(6) - 40 = -36 + 72 - 40 = -4$.
5. Vertex is $(6, -4)$.
6. Equation: $y = -(x - 6)^2 - 4$.
Problem 13: $y = x^2 + 10x + 33$
1. $a=1, b=10, c=33$.
2. $h = -10 / 2 = -5$.
3. $k = (-5)^2 + 10(-5) + 33 = 25 - 50 + 33 = 8$.
4. Vertex is $(-5, 8)$.
5. Equation: $y = (x + 5)^2 + 8$.
Problem 14: $y = 5(x + 2)^2$
1. This is already in vertex form $y = a(x-h)^2 + k$.
2. $h = -2, k = 0$.
3. Equation: $y = 5(x + 2)^2 + 0$ or just $y = 5(x + 2)^2$.
Final Answer:
1) $y = (x + 8)^2 + 7$
2) $y = (x - 1)^2 - 6$
3) $y = (x - 7)^2 + 10$
4) $y = (x + 4.5)^2 + 149.75$
5) $y = (x - 6)^2 + 10$
6) $y = (x + 2)^2 - 4$
7) $y = (x - 3)^2 - 4$
8) $y = (x + 4.5)^2 - 0.25$
9) $x = 2(y - 2)^2 - 4$
10) $y = -6(x + 1)^2 - 7$
11) $x = -\frac{1}{18}(y + 4)^2 + \frac{8}{9}$
12) $y = -(x - 6)^2 - 4$
13) $y = (x + 5)^2 + 8$
14) $y = 5(x + 2)^2$
The task is to convert equations of parabolas into Vertex Form.
The standard vertex form of a parabola is:
$$y = a(x - h)^2 + k$$
where $(h, k)$ is the vertex.
I will solve each of the 14 problems step-by-step.
Problem 1: $y = x^2 + 16x + 71$
1. Identify $a=1$, $b=16$, $c=71$.
2. Find the x-coordinate of the vertex: $h = -b / 2a = -16 / 2 = -8$.
3. Find the y-coordinate by plugging $h$ back in: $k = (-8)^2 + 16(-8) + 71 = 64 - 128 + 71 = 7$.
4. Vertex is $(-8, 7)$.
5. Equation: $y = (x - (-8))^2 + 7 \rightarrow y = (x + 8)^2 + 7$.
Problem 2: $y = x^2 - 2x - 5$
1. $a=1, b=-2, c=-5$.
2. $h = -(-2) / 2 = 1$.
3. $k = (1)^2 - 2(1) - 5 = 1 - 2 - 5 = -6$.
4. Vertex is $(1, -6)$.
5. Equation: $y = (x - 1)^2 - 6$.
Problem 3: $y = x^2 - 14x + 59$
1. $a=1, b=-14, c=59$.
2. $h = -(-14) / 2 = 7$.
3. $k = (7)^2 - 14(7) + 59 = 49 - 98 + 59 = 10$.
4. Vertex is $(7, 10)$.
5. Equation: $y = (x - 7)^2 + 10$.
Problem 4: $y = x^2 + 9x + 170$
1. $a=1, b=9, c=170$.
2. $h = -9 / 2 = -4.5$.
3. $k = (-4.5)^2 + 9(-4.5) + 170 = 20.25 - 40.5 + 170 = 149.75$.
4. Vertex is $(-4.5, 149.75)$.
5. Equation: $y = (x + 4.5)^2 + 149.75$.
Problem 5: $y = x^2 - 12x + 46$
1. $a=1, b=-12, c=46$.
2. $h = -(-12) / 2 = 6$.
3. $k = (6)^2 - 12(6) + 46 = 36 - 72 + 46 = 10$.
4. Vertex is $(6, 10)$.
5. Equation: $y = (x - 6)^2 + 10$.
Problem 6: $y = x^2 + 4x$
1. $a=1, b=4, c=0$.
2. $h = -4 / 2 = -2$.
3. $k = (-2)^2 + 4(-2) = 4 - 8 = -4$.
4. Vertex is $(-2, -4)$.
5. Equation: $y = (x + 2)^2 - 4$.
Problem 7: $y = x^2 - 6x + 5$
1. $a=1, b=-6, c=5$.
2. $h = -(-6) / 2 = 3$.
3. $k = (3)^2 - 6(3) + 5 = 9 - 18 + 5 = -4$.
4. Vertex is $(3, -4)$.
5. Equation: $y = (x - 3)^2 - 4$.
Problem 8: $y = (x + 5)(x + 4)$
1. First expand to standard form: $y = x^2 + 4x + 5x + 20 \rightarrow y = x^2 + 9x + 20$.
2. $a=1, b=9, c=20$.
3. $h = -9 / 2 = -4.5$.
4. $k = (-4.5)^2 + 9(-4.5) + 20 = 20.25 - 40.5 + 20 = -0.25$.
5. Vertex is $(-4.5, -0.25)$.
6. Equation: $y = (x + 4.5)^2 - 0.25$.
Problem 9: $\frac{1}{2}(x + 4) = (y - 2)^2$
1. This is a horizontal parabola. We need to isolate $y$.
2. Multiply by 2: $x + 4 = 2(y - 2)^2$.
3. Divide by 2: $\frac{1}{2}(x + 4) = (y - 2)^2$.
4. Take square root? No, usually vertex form for horizontal is $x = a(y-k)^2 + h$.
5. Let's rearrange to $x = \dots$
6. $x + 4 = 2(y-2)^2 \rightarrow x = 2(y-2)^2 - 4$.
7. Vertex is $(-4, 2)$.
8. Equation: $x = 2(y - 2)^2 - 4$.
Problem 10: $6x^2 + 12x + y + 13 = 0$
1. Isolate $y$: $y = -6x^2 - 12x - 13$.
2. $a=-6, b=-12, c=-13$.
3. $h = -(-12) / (2 \cdot -6) = 12 / -12 = -1$.
4. $k = -6(-1)^2 - 12(-1) - 13 = -6 + 12 - 13 = -7$.
5. Vertex is $(-1, -7)$.
6. Equation: $y = -6(x + 1)^2 - 7$.
Problem 11: $162x + 72y = -9y^2$
1. Rearrange to solve for $x$ (since $y^2$ term exists).
2. $162x = -9y^2 - 72y$.
3. Divide by 162: $x = \frac{-9}{162}y^2 - \frac{72}{162}y$.
4. Simplify fractions: $x = -\frac{1}{18}y^2 - \frac{4}{9}y$.
5. Find vertex y-coordinate: $k = -b / 2a = -(-4/9) / (2 \cdot -1/18) = (4/9) / (-1/9) = -4$.
6. Find vertex x-coordinate: Plug $y=-4$ into equation.
$x = -\frac{1}{18}(-4)^2 - \frac{4}{9}(-4) = -\frac{16}{18} + \frac{16}{9} = -\frac{8}{9} + \frac{16}{9} = \frac{8}{9}$.
7. Vertex is $(8/9, -4)$.
8. Equation: $x = -\frac{1}{18}(y + 4)^2 + \frac{8}{9}$.
Problem 12: $x^2 - 12x + y + 40 = 0$
1. Isolate $y$: $y = -x^2 + 12x - 40$.
2. $a=-1, b=12, c=-40$.
3. $h = -12 / (2 \cdot -1) = 6$.
4. $k = -(6)^2 + 12(6) - 40 = -36 + 72 - 40 = -4$.
5. Vertex is $(6, -4)$.
6. Equation: $y = -(x - 6)^2 - 4$.
Problem 13: $y = x^2 + 10x + 33$
1. $a=1, b=10, c=33$.
2. $h = -10 / 2 = -5$.
3. $k = (-5)^2 + 10(-5) + 33 = 25 - 50 + 33 = 8$.
4. Vertex is $(-5, 8)$.
5. Equation: $y = (x + 5)^2 + 8$.
Problem 14: $y = 5(x + 2)^2$
1. This is already in vertex form $y = a(x-h)^2 + k$.
2. $h = -2, k = 0$.
3. Equation: $y = 5(x + 2)^2 + 0$ or just $y = 5(x + 2)^2$.
Final Answer:
1) $y = (x + 8)^2 + 7$
2) $y = (x - 1)^2 - 6$
3) $y = (x - 7)^2 + 10$
4) $y = (x + 4.5)^2 + 149.75$
5) $y = (x - 6)^2 + 10$
6) $y = (x + 2)^2 - 4$
7) $y = (x - 3)^2 - 4$
8) $y = (x + 4.5)^2 - 0.25$
9) $x = 2(y - 2)^2 - 4$
10) $y = -6(x + 1)^2 - 7$
11) $x = -\frac{1}{18}(y + 4)^2 + \frac{8}{9}$
12) $y = -(x - 6)^2 - 4$
13) $y = (x + 5)^2 + 8$
14) $y = 5(x + 2)^2$
Parent Tip: Review the logic above to help your child master the concept of converting quadratic equations worksheet standard to vertex.