Problem 1 :
Find the quadratic function in vertex form, if the vertex is (2, 3). and its graph is passing through the point (5, 21).
Problem 2 :
Find the quadratic function in standard form, if the vertex is (-1, -2). and its graph is passing through the point (0, 4).
Problem 3 :
Find the quadratic function in standard form, if the two x-intercepts are 2 and 5 and its graph is passing through the point (1, 3).
Problem 4 :
Find the quadratic function in standard form, if the vrtex is (5, 7) and its graph is passing through the point (-1, 0).
Problem 5 :
Find the quadratic function in vertex form, if the maximum point is (2, 5) and x-intercept is 1.
Problem 6 :
Find the quadratic function in vertex form, if the axis of symmetry is x = 5, range yis and y-intercept is
1. Answer :
Quadratic function in vertex form :
y = a(x - h)^{2} + k
Here (h, k) is the vertex.
Given : Vertex is (2, 3).
Substitute (h, k) = (2, 3).
y = a(x - 2)^{2} + 3
Given : Graph is passing through the point (5, 21).
Substitute x = 5 and y = 21.
21 = a(5 - 2)^{2} + 3
21 = a(3)^{2} + 3
21 = 9a + 3
Subtract 3 from both sides.
18 = 9a
Divide both sides by 9.
2 = a
The required quadratic function in vertex form :
y = 2(x - 2)^{2} + 3
2. Answer :
Quadratic function in vertex form :
y = a(x - h)^{2} + k
Here (h, k) is the vertex.
Given : Vertex is (-1, -2).
Substitute (h, k) = (-1, -2).
y = a[x - (-1)]^{2} + (-2)
y = a(x + 1)^{2} - 2
Given : Graph is passing through the point (0, 4).
Substitute x = 0 and y = 4.
4 = a(0 + 1)^{2} - 2
4 = a(1)^{2} - 2
4 = a - 2
Add 2 to both sides.
6 = a
The required quadratic function in vertex form :
y = 6(x + 1)^{2} - 2
3. Answer :
Quadratic function in standard form :
y = ax^{2} + bx + c
If the x = p and y = q are two x-intercepts, then the quadratic function is
y = a(x - p)(x - q)
Since x = 2 and x = 5 are two x-intercepts, the quadratic function is
y = a(x - 2)(x - 5)
Given : Graph is passing through the point (1, 8).
Substitute x = 1 and y = 8.
8 = a(1 - 2)(1 - 5)
8 = a(-1)(-4)
8 = 4a
Divide both sies by 4.
2 = a
The required quadratic function in standard form :
y = 2(x - 2)(x - 5)
y = 2(x^{2} - 5x - 2x + 10)
y = 2(x^{2} - 7x + 10)
y = 2x^{2} - 14x + 20
4. Answer :
Since the vertex is given, initially we can write the quadratic function in vertex form and convert it to standard form.
Quadratic function in vertex form :
y = a(x - h)^{2} + k
Given : Vertex is (5, 7).
Substitute (h, k) = (5, 7).
y = a(x - 5)^{2} + 7
Given : Graph is passing through the point (4, 0).
Substitute x = 4 and y = 0.
0 = a(4 - 5)^{2} + 7
0 = a(-1)^{2} + 7
0 = a(1) + 7
0 = a + 7
Subtract 7 from both sides.
-7 = a
The required quadratic function in vertex form :
y = -7(x - 5)^{2} + 7
Convert it to standard form.
y = -7(x^{2} - 10x + 25) + 7
y = -7x^{2} + 70x - 175 + 7
y = -7x^{2} + 70x - 168
5. Answer :
The graph of a quadratic function is a parabola.
Sketch a parabola with the maximum point (2, 5) and x-intercept (1, 0).
The vertex of the parabola above is (2, 5).
Quadratic function in vertex form :
y = a(x - h)^{2} + k
Substitute (h, k) = (2, 5).
y = a(x - 2)^{2} + 5
Since the x-intercept is (1, 0), substitute x = 1 and y = 0 to find the value of a.
0 = a(1 - 2)^{2} + 5
0 = a(-1)^{2} + 5
0 = a(1) + 5
0 = a + 5
-5 = a
The required quadratic function in vertex form :
y = -5(x - 2)^{2} + 5
6. Answer :
Sketch a parabola for the following :
Axis of symmetry : x = 5
Range : y ≥ 2
y-intercept : (0, 7)
The vertex of the parabola above is (5, 2).
Quadratic function in vertex form :
y = a(x - h)^{2} + k
Substitute (h, k) = (5, 2).
y = a(x - 5)^{2} + 2
Since the y-intercept is (0, 7), substitute x = 0 and y = 7 to find the value of a.
7 = a(0 - 5)^{2} + 2
7 = a(-5)^{2} + 2
7 = 25a + 2
5 = 25a
⅕ = a
The required quadratic function in vertex form :
Convert it to standard form.
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