IGCSE Mathematics: How to find the quadratic functions with information given

IGCSE Mathematics: How to find quadratic functions with given information.

IGCSE may provide you with:

Vertex and another point
x-intercepts and a point
Vertex or intercepts with a = 1
Remember: y = a(x – h)² + k – this is known as the vertex form of a quadratic function.
a is the constant.
(h, k) is the coordinate of the vertex.
Vertex forms have their uses, as you should be aware of:

If a < 0, we have a function with a maximum.
If a > 0, we have a function with a minimum.

Important to know:
When the vertex form is y = a(x + h)² + k,
the coordinate of the vertex will be (-h, k) since the value of h must be negative for the original vertex form to change: (x – (-h)) + k = (x + h).

Likewise,
When the vertex form is y = a(x – h)² – k,
the coordinate of the vertex will be (h, -k) since the value of k must be negative for the original vertex form to change: a(x – h)² + (-k) = a(x – h)² – k.

GIVEN VERTEX AND ANOTHER POINT:
Find the equation of the quadratic equation given the vertex (2, -4) and the y-intercept of 8.
y = a(x – h)² + k

1st piece of information given: vertex (2, -4), therefore, h = 2 and k = -4.
Substitute,
y = a(x – 2)² – 4.

2nd piece of information given: y-intercept of 8, therefore, x = 0 and y = 8.
Substitute,
8 = a(0 – 2)² – 4
8 = a(-2)² – 4
8 = 4a – 4
8 + 4 = 4a – 4 + 4
4a = 12
a = 3

Therefore, the equation is y = 3(x – 2)² – 4 or 3x² – 12x + 8.

Get quality help now

Doctor Jennifer

Verified

Proficient in: Mathematics

5 (893)

“ Thank you so much for accepting my assignment the night before it was due. I look forward to working with you moving forward ”

+84 relevant experts are online

Hire writer

GIVEN X-INTERCEPT AND A POINT:
Find the quadratic equation with x-intercepts (-2, 0) and (2, 0) and passing through the point (1, 3).
y = ax² + bx + c

For this question, you have to find a, b, and c.

1st piece of information given: x = -2, y = 0, and x = 2, y = 0.
Substitute:
0 = 4a – 2b + c [1]
0 = 4a + 2b + c [2]
[1] – [2] = 0 = -4b

Therefore, b = 0.

2nd piece of information given: x = 1 and y = 3.
Substitute:
3 = a + b + c [Since b = 0]
c = 3 – a [3]
Substitute [3] into [1]:
0 = 4a + 3 – a

3a + 3 = 0

Therefore, a = -1.

Since a = -1 and b = 0,
We can substitute this information into [1] to find c:
0 = -4 + c

Hence, c = 4.

Consider y = ax² + bx + c,
y = -x² + 4.

VERTEX OR INTERCEPTS WITH A = 1:
Find the quadratic equation in the form y = a(x – b)² + c,
when the vertex is at (3, -1) and the value of a = 1.

Information given: a = 1, h = 3, and k = -1."