Question 1 - Vertex form — finding a from vertex and y-intercept conditions Practice | Prasnya
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Question 1
00:00
Est. 3 min
Marks:
+4.00
0.00
Consider the two quadratic functions
y
1
=
x
2
−
4
x
+
1
y_1 = x^2 - 4x + 1
and
y
2
=
x
2
−
6
x
−
3
y_2 = x^2 - 6x - 3
. A quadratic function
f
(
x
)
=
a
x
2
+
b
x
+
c
f(x) = ax^2 + bx + c
has its vertex at the intersection point of
y
1
y_1
and
y
2
y_2
, its
y
y
-intercept is 2, and its axis of symmetry is parallel to the
y
y
-axis. What is the equation of
f
(
x
)
f(x)
?
Single correct
Medium Difficulty
26 October 2025
A
y
=
(
−
11
4
)
(
x
+
2
)
2
+
13
y = \left(-\dfrac{11}{4}\right)(x+2)^2 + 13
B
y
=
(
−
11
4
)
(
x
−
2
)
2
+
13
y = \left(-\dfrac{11}{4}\right)(x-2)^2 + 13
C
y
=
(
11
4
)
(
x
+
2
)
2
+
13
y = \left(\dfrac{11}{4}\right)(x+2)^2 + 13
D
y
=
(
−
11
4
)
(
x
+
2
)
2
−
13
y = \left(-\dfrac{11}{4}\right)(x+2)^2 - 13
