Graphing Quadratics — Challenge
Every parabola has a vertex, an axis of symmetry, and up to two roots. Find those and the graph draws itself.
What these problems cover
These questions use 9 ideas from the course. Follow any of them for a definition, where it is taught, and what it leads to.
- Open — Open excludes the endpoint (<, >); closed includes it (≤, ≥).
- Factor — A number that divides evenly into another.
- y-intercept — The output when x = 0 — where the line crosses the vertical axis.
- Parabola — The U-shaped graph of a quadratic function.
- Vertex — The turning point — the minimum if it opens up, the maximum if it opens down.
- Axis of symmetry — The vertical line through the vertex, x = -b2a.
- Roots — The x values where y = 0.
- Vertex form — y = a(x - h)2 + k, with vertex at (h, k).
- Height — The perpendicular distance from the base to the opposite corner.
Problem 1
What is the vertex of y=x2?
- (0,0)
- (1,1)
- (0,1)
- (1,0)
- (-1,0)
Worked answer
(0,0)
The graph of y = x2 is a parabola opening upwards, and its lowest point is its vertex.
Since x2 is never negative and equals 0 only when x=0, the lowest point is at x=0, where y=0.
So the vertex is the origin, (0,0).
Problem 2
What is the vertex of y=(x-3)2+2?
- (-3,2)
- (3,-2)
- (3,2)
- (2,3)
- (-3,-2)
Worked answer
(3,2)
In the form y = (x-h)2+k the vertex is at (h,k).
Here h = 3, because the bracket is x-3, and k = 2.
So the vertex is (3,2). The sign inside the bracket is the opposite of the x coordinate, which is the usual trap.
Problem 3
What is the axis of symmetry of y=x2-4x+1?
- x=-2
- x=2
- x=4
- x=1
- x=-4
Worked answer
x=2
The axis of symmetry is the vertical line x = -b2a.
Here a=1 and b=-4, so x = --42 = 2.
The parabola is symmetric about x=2, and its vertex lies on that line.
Problem 4
Does the parabola y=-x2+3 open upwards or downwards?
- Upwards
- Downwards
- Neither
- It is a straight line
- It depends on x
Worked answer
Downwards
The direction depends on the sign of the x2 coefficient.
Here that coefficient is -1, which is negative, so the parabola opens downwards.
Its vertex at (0,3) is therefore a maximum rather than a minimum.
Problem 5
What is the y-intercept of y=x2-2x-3?
- 3
- -3
- -2
- 2
- 0
Worked answer
-3
The y-intercept is the value of y when x=0.
y = 0-0-3 = -3.
In a quadratic written this way, the constant term is always the y-intercept.
Problem 6
What is the positive x-intercept of y=x2-4?
- 1
- 2
- 4
- -2
- 16
Worked answer
2
The x-intercepts are where y=0, so solve x2-4 = 0.
That gives x2 = 4, so x = 2 or x = -2.
The positive one is 2. The parabola crosses the axis twice, symmetrically about x=0.
Problem 7
What is the vertex of y=x2+6x+5?
- (3,-4)
- (-3,4)
- (-3,-4)
- (-6,5)
- (3,4)
Worked answer
(-3,-4)
The vertex lies on the axis of symmetry, at x = -b2a = -62 = -3.
Substitute to find y: (-3)2+6(-3)+5 = 9-18+5 = -4.
So the vertex is (-3,-4).
Problem 8
How many x-intercepts does y=x2+1 have?
- 0
- 1
- 2
- 3
- Infinitely many
Worked answer
0
An x-intercept needs x2+1 = 0, that is x2 = -1.
No real number squares to a negative, so there is no real solution.
The parabola opens upwards with its lowest point at (0,1), entirely above the x-axis.