A parabola is the smooth curve made by a quadratic graph, and it has a vertical axis of symmetry. Its highest or lowest point is called the turning point , or sometimes the vertex .

National 5MathsSQAFlashcardsAlgebraic SkillsQuadratic Functions

Quadratic Functions (SQA National 5 Maths): Flashcards

Exam code: X847 75

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FrontFeatures of Quadratic Graphs
Define a parabola.

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Cards in this collection (23)

Define a parabola.
A parabola is the smooth curve made by a quadratic graph, and it has a vertical axis of symmetry.
Its highest or lowest point is called the turning point, or sometimes the vertex.
What decides whether a quadratic graph opens upwards or downwards?
The sign of the number in front of $x^{2}$ decides it, so a positive one gives a $\cup$ shape and a negative one a $\cap$ shape.
A $\cup$ shape has a minimum turning point, and a $\cap$ shape has a maximum one.
True or False?
A quadratic graph always crosses the $y$ axis.
True.
Every value of $x$ gives a value of $y$ , including $x = 0$ , so there is always exactly one crossing point on the $y$ axis.
Substituting $x = 0$ into the equation gives the value of $y$ at that point.
For $y = {(x + a)}^{2} + b$ where does the turning point lie?
The turning point is at $(- a, b)$ , so the $x$ coordinate has its sign swapped while the $y$ coordinate does not.
That holds whether the turning point is a maximum or a minimum.
Complete the coordinates of the turning point of this quadratic.
$y = {(x + 5)}^{2} - 18 has turning point (______,______)$
The completed coordinates are:
$(- 5, - 18)$
The $5$ inside the bracket becomes $- 5$ , while the $- 18$ carries straight through unchanged.
How do you write down the axis of symmetry once you know the turning point?
The axis of symmetry is the vertical line through the turning point, so its equation uses the turning point's $x$ coordinate.
For a turning point at $(3, 7)$ the axis of symmetry is $x = 3$ .
How does symmetry help you find a second point at the same height as a known one?
Two points on a parabola with the same $y$ value sit the same distance from the axis of symmetry, one on each side.
If the axis is $x = - 5$ and one point is at $x = 0$ , which is $5$ to the right, the other is $5$ to the left at $x = - 10$ .
What form does a quadratic take when its turning point is at the origin?
Its equation is $y = k x^{2}$ , with the $y$ axis as its axis of symmetry.
Only the value of $k$ is then left to find.
How is the multiplier $k$ found once the rest of a quadratic's equation is known?
Substitute the coordinates of any other point on the graph and solve the resulting equation.
For a graph through the origin and $(2, - 12)$ that gives $- 12 = k {(2)}^{2}$ , so $4 k = - 12$ and $k = - 3$ .
How do you get $a$ and $b$ in $y = k {(x + a)}^{2} + b$ from a graph?
Read the turning point off the graph, since it sits at $(- a, b)$ .
Remember to switch the sign of the $x$ coordinate to get $a$ , so a turning point at $(- 2, 3)$ gives $a = 2$ and $b = 3$ .
True or False?
In $y = k {(x + a)}^{2} + b$ the graph crosses the $y$ axis at $(0, b)$ .
False.
The $b$ is the $y$ coordinate of the turning point, not of the $y$ axis crossing.
For $y = 2 {(x + 2)}^{2} + 3$ the turning point is at $(- 2, 3)$ but the curve crosses the $y$ axis at $11$ , whereas $y = a x^{2} + b x + c$ really does cross at $(0, c)$ .
A parabola $y = {(x + a)}^{2} + b$ has its turning point at $(4, 3)$ so complete the two values.
$a =______and b =______$
The completed values are:
$a = - 4 and b = 3$
The turning point sits at $(- a, b)$ , so $- a = 4$ gives $a = - 4$ while $b$ is read off directly.
Once the equation is known how do you find where the curve crosses the $y$ axis?
Substitute $x = 0$ into the equation and work out the value of $y$ .
For $y = {(x - 4)}^{2} + 3$ that gives ${(0 - 4)}^{2} + 3 = 16 + 3 = 19$ .
What is the least you need from a graph to pin down its quadratic equation?
The turning point and one other point on the curve are enough.
The turning point fixes $a$ and $b$ , and the other point then fixes $k$ .
What is the connection between the solutions of $a x^{2} + b x + c = 0$ and its graph?
Each solution $x = p$ gives a point where the graph of $y = a x^{2} + b x + c$ crosses the $x$ axis, at $(p, 0)$ .
That works because $y = 0$ is the equation of the $x$ axis.
What does the number of solutions of a quadratic equation say about its graph?
Two solutions means the graph crosses the $x$ axis at two points, and one solution means it touches the axis at a single point.
No solutions means it misses the $x$ axis altogether, lying wholly above it or wholly below it.
From $y = (2 x - 1) (2 x - 7)$ how do you find the $x$ axis intercepts?
Set each bracket equal to zero and solve, so $2 x - 1 = 0$ gives $x = 0.5$ and $2 x - 7 = 0$ gives $x = 3.5$ .
The intercepts are therefore at $(0.5, 0)$ and $(3.5, 0)$ .
How do you find the axis of symmetry from the two $x$ axis intercepts?
The axis of symmetry lies midway between them, so take the average of the two $x$ values.
For intercepts at $0.5$ and $3.5$ that gives $\frac{0.5 + 3.5}{2} = 2$ , so the axis of symmetry is $x = 2$ .
Once you know the axis of symmetry how do you get the turning point?
Substitute that $x$ value into the equation to find the matching $y$ value.
For $y = (2 x - 1) (2 x - 7)$ at $x = 2$ that gives $(3) \times (- 3) = - 9$ , so the turning point is $(2, - 9)$ .
Complete the turning point of this quadratic given in completed square form.
$y = 2 {(x - 4)}^{2} + 1 has turning point (______,______)$
The completed turning point is:
$(4, 1)$
The number inside the bracket has its sign swapped, so $(x - 4)$ gives $4$ , while the $+ 1$ outside carries through unchanged.
In $y = k {(x + p)}^{2} + q$ what does the $k$ tell you?
The sign of $k$ gives the shape, with a positive $k$ making a $\cup$ shape and a negative $k$ a $\cap$ shape.
So $y = 2 {(x - 4)}^{2} + 1$ opens upwards, while $y = - {(x + 2)}^{2} - 1$ opens downwards.
How are the $x$ axis intercepts of $y = 2 {(x - 4)}^{2} - 18$ worked out?
Set the whole expression equal to zero and solve for $x$ .
That gives ${(x - 4)}^{2} = 9$ , then $x - 4 = \pm 3$ , so the intercepts are at $(1, 0)$ and $(7, 0)$ .
True or False?
A quadratic graph that never crosses the $x$ axis still has a turning point and a crossing on the $y$ axis.
True.
Every parabola has a turning point and meets the $y$ axis once, whatever it does about the $x$ axis.
The graph of $y = - {(x + 2)}^{2} - 1$ never reaches the $x$ axis, but its turning point is $(- 2, - 1)$ and it crosses the $y$ axis at $(0, - 5)$ .

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