To find equation of parabola from the given vertex and a point, we may use vertex form of the parabola.
y = a(x - h)^{2} + k
Here (h, k) is the vertex.
To find the value of a, we can apply the given point into the equation instead of (x, y).
Problem 1 :
A parabola with vertex (1, 2) ; passes through (2, 3)
Solution :
The vertex is (1, 2).
So, the equation is f(x) = a(x – 1)^{2} + 2. Find a by substituting the coordinates of another point on the parabola, such as (2, 3).
3 = a(2 – 1)^{2} + 2
3 = a(1)^{2} + 2
3 = a + 2
Substituting 2 on both sides.
3 – 2 = a + 2 – 2
1 = a
The equation for the parabola is f(x) = (x – 1)^{2} + 2.
Use the description of the parabola to write its equation in vertex form.
Problem 2 :
A parabola with vertex (0, -3) ; passes through (3, 0)
Solution :
The vertex is (0, -3), so, the equation is f(x) = a(x – 0)^{2} - 3. Find a by substituting the coordinates of another point on the parabola, such as (3, 0).
0 = a(3 – 0)^{2} - 3
0 = a(3)^{2} - 3
0 = 9a - 3
9a = 3
Dividing 9 on both sides.
9a/9 = 3/9
a = 1/3
f(x) = (1/3)(x – 0)^{2} - 3
The equation for the parabola is f(x) = (1/3)x^{2} - 3.
Problem 3 :
A parabola with vertex (-1, 4) ; passes through (-2, 3)
Solution :
The vertex is (-1, 4), so, the equation is f(x) = a(x + 1)^{2} + 4. Find a by substituting the coordinates of another point on the parabola, such as (-2, 3).
3 = a(-2 + 1)^{2} + 4
3 = a(-1)^{2} + 4
3 = a + 4
Substituting 4 on both sides.
3 – 4 = a + 4 – 4
-1 = a
The equation for the parabola is f(x) = -1(x + 1)^{2} + 4.
For each graph, write a quadratic function in vertex form and standard form.
Problem 4 :
Solution :
Quadratic function in vertex form :
y = a(x – h)^{2} + k
Vertex v = (h, k)
h = -1, k = -2
(h, k) = (-1, -2)
(x, y) = (3, 2)
y = a(x + 1)^{2} – 2
2 = a(3 + 1)^{2} – 2
2 = a(4)^{2} – 2
2 = 16a – 2
Adding 2 on both sides.
2 + 2 = 16a – 2 + 2
4 = 16a
Dividing 4 on both sides.
4/4 = 16a/4
1 = 4a
1/4 = a
Vertex form :
y = (1/4)(x + 1)^{2} – 2
Standard form :
y = 1/4(x^{2} + 2x + 1) – 2
y = 1/4x^{2} + 2/4x + 1/4 – 2
y = (x^{2} + 2x + 1 – 8)/4
y = (1/4) (x^{2} + 2x – 7)
Problem 5 :
Solution :
Quadratic function in vertex form :
y = a(x – h)^{2} + k
Vertex v = (h, k)
h = 2, k = 3
(h, k) = (2, 3)
(x, y) = (4, 7)
y = a(x - 2)^{2} + 3
7 = a(4 - 2)^{2} + 3
7 = a(2)^{2} + 3
7 = 4a + 3
Subtracting 3 on both sides.
7 – 3 = 4a + 3 – 3
4 = 4a
Dividing 4 on both sides.
4/4 = 4a/4
1 = a
Vertex form :
y = (x - 2)^{2} + 3
Standard form :
y = x^{2} – 2(x)(2) + 2^{2 }+ 3
= x^{2} – 4x + 4 + 3
= x^{2} – 4x + 7
Problem 6 :
What is the equation for the parabola shown in the graph ?
Solution :
The vertex is (2, 3), so, the equation is f(x) = a(x – 2)^{2} + 3. Find a by substituting the coordinates of another point on the parabola, such as (4, -1).
-1 = a(4 – 2)^{2} + 3
-1 = a(2)^{2} + 3
-4 = 4a
a = -1
The equation for the parabola is f(x) = -(x – 2)^{2} + 3.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
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