A parabola is the curve formed by any quadratic graph. It is a "U" shape, which may open upwards or downwards.

True or False? Every parabola has exactly one turning point.

True. A quadratic graph turns once and only once, at its maximum or minimum point. It is the only place where the curve changes direction.

What should a sketch of a quadratic graph show?

The shape and orientation of the parabola, the axes intercepts , and the coordinates of the turning point . All of them can be worked out from the equation.

Of the three standard methods for solving a quadratic equation (factorising, completing the square and the quadratic formula), which of them always work?

Completing the square and the quadratic formula both work for every quadratic. Factorising is the quickest when it works, but not every quadratic will factorise.

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Quadratics (Edexcel A Level Maths: Pure): Flashcards

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FrontQuadratic Graphs
Define parabola.

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Cards in this collection (30)

Define parabola.
A parabola is the curve formed by any quadratic graph.
It is a "U" shape, which may open upwards or downwards.
On the graph of $y = a x^{2} + b x + c$ , how does the sign of $a$ change the shape?
If $a > 0$ the parabola is upright, $\cup$ , and has a minimum point.
If $a < 0$ it is upside down, $\cap$ , and has a maximum point.
For the graph of $y = a x^{2} + b x + c$ , the $y$ -axis intercept is the point $(0,______)$ .
For the graph of $y = a x^{2} + b x + c$ , the $y$ -axis intercept is the point $(0, c)$ .
Substituting $x = 0$ removes both of the other terms, leaving only the constant.
What do the roots of a quadratic function tell you about its graph?
They are the values of $x$ where the graph crosses the $x$ -axis.
You find them by setting $y = 0$ and solving the equation that results.
True or False?
Every parabola has exactly one turning point.
True.
A quadratic graph turns once and only once, at its maximum or minimum point.
It is the only place where the curve changes direction.
What should a sketch of a quadratic graph show?
The shape and orientation of the parabola, the axes intercepts, and the coordinates of the turning point.
All of them can be worked out from the equation.
Why might you need to rearrange a quadratic equation before sketching its graph?
Because the coefficients are read off the standard form $y = a x^{2} + b x + c$ , and an equation such as $3 x^{2} = y - 5 x + 12$ hides them.
Until it is rearranged you cannot tell which coefficient is which.
Define discriminant.
The discriminant of $a x^{2} + b x + c = 0$ is $b^{2} - 4 a c$ , the expression under the square root in the quadratic formula.
It is sometimes written as $Δ$ .
What do the three possible signs of the discriminant tell you about the roots of a quadratic equation?
They give the number of real roots:
- $b^{2} - 4 a c > 0$ : two distinct real roots
- $b^{2} - 4 a c = 0$ : one real root, also called a repeated root
- $b^{2} - 4 a c < 0$ : no real roots
True or False?
A quadratic whose discriminant is zero never meets the $x$ -axis.
False.
A zero discriminant means the graph touches the $x$ -axis at exactly one point.
It is a negative discriminant that means the graph never meets the axis at all.
Why does a discriminant of zero give a repeated root rather than two separate ones?
Because the quadratic formula adds and subtracts $\sqrt{b^{2} - 4 a c}$ , and adding or subtracting zero gives the same value both times.
The two roots therefore coincide.
When a question asks only for real roots of a quadratic, the condition to use is $b^{2} - 4 a c______0$ .
When a question asks only for real roots of a quadratic, the condition to use is $b^{2} - 4 a c \geq 0$ .
"Real" covers distinct and repeated roots alike, so the case of zero has to be included.
A quadratic in $x$ has coefficients involving an unknown constant $k$ . How do you find the values of $k$ that give two distinct real roots?
Write the discriminant in terms of $k$ and set it greater than zero, which produces an inequality in $k$ to solve.
For $k x^{2} + 3 k x - 2 k^{2} = 0$ this gives ${(3 k)}^{2} - 4 (k) (- 2 k^{2}) > 0$ , that is $9 k^{2} + 8 k^{3} > 0$ .
Define completing the square.
Completing the square means writing $y = a x^{2} + b x + c$ in the form $y = a {(x + p)}^{2} + q$ .
Every $x$ then sits inside a single bracket.
When $a = 1$ , completing the square on $y = x^{2} + b x + c$ uses $p =______$ and $q =______$ .
When $a = 1$ , completing the square on $y = x^{2} + b x + c$ uses $p = \frac{b}{2}$ and $q = c - p^{2}$ .
For $y = x^{2} + 8 x - 2$ this gives $p = 4$ and $q = - 2 - 4^{2} = - 18$ , so $y = {(x + 4)}^{2} - 18$ .
What is the first thing to do when completing the square and $a \neq 1$ ?
Factorise $a$ out of the $x^{2}$ and $x$ terms only, leaving the constant outside the bracket.
So $y = 4 x^{2} + 16 x + 5$ becomes $y = 4 [x^{2} + 4 x] + 5$ , and the work continues inside the bracket.
True or False?
Completing the square on $y = 4 x^{2} + 16 x + 5$ gives $y = 4 {(x + 2)}^{2} + 5$ .
False.
The constant changes. Inside the bracket the working gives ${(x + 2)}^{2} - 4$ , and the factor of $4$ turns that $- 4$ into $- 16$ .
The correct form is $y = 4 {(x + 2)}^{2} - 11$ .
A quadratic is written as $y = a {(x + p)}^{2} + q$ . Where is its turning point?
The turning point is at $(- p, q)$ , whether it is a maximum or a minimum.
The link works both ways: given a turning point you can write the quadratic down in this form straight away, then use one other point on the curve to find $a$ .
How does the completed square form show that $y = x^{2} + 6 x - 3$ is never less than $- 12$ ?
Completing the square gives $y = {(x + 3)}^{2} - 12$ , and a squared term is never negative.
The smallest value ${(x + 3)}^{2}$ can take is $0$ , so the smallest value of $y$ is $- 12$ .
Of the three standard methods for solving a quadratic equation (factorising, completing the square and the quadratic formula), which of them always work?
Completing the square and the quadratic formula both work for every quadratic.
Factorising is the quickest when it works, but not every quadratic will factorise.
Complete the quadratic formula for $a x^{2} + b x + c = 0$ :
$x = \frac{- b \pm \sqrt{______}}{______}$
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
The $2 a$ divides all of the top line, not just the square root.
After factorising a quadratic to $(x + 11) (x - 2) = 0$ , why can you set each bracket equal to zero?
Because if a product of two numbers is zero, at least one of them must be zero.
So $x + 11 = 0$ or $x - 2 = 0$ , giving $x = - 11$ or $x = 2$ .
True or False?
The equation $x^{2} = 9$ has only the solution $x = 3$ .
False.
Both $3$ and $- 3$ square to give $9$ , so the solutions are $x = 3$ and $x = - 3$ .
Solving $x^{2} = 9$ is what produces two roots; the symbol $\sqrt{9}$ on its own means the positive root only.
The equation $x + 5 = \frac{12}{x}$ does not look like a quadratic. What do you do first?
Multiply every term by $x$ to clear the fraction, then collect everything on one side.
That gives $x^{2} + 5 x = 12$ , and then $x^{2} + 5 x - 12 = 0$ , which is the form every method needs.
Once an equation has been written as ${(x + 5)}^{2} - 49 = 0$ , how do you finish solving it?
Move the constant across, then take the square root of both sides: ${(x + 5)}^{2} = 49$ , so $x + 5 = \pm 7$ .
That gives $x = 2$ or $x = - 12$ .
Define hidden quadratic.
A hidden quadratic is an equation of the form $a {[f (x)]}^{2} + b f (x) + c = 0$ .
It is an ordinary quadratic with $x$ replaced throughout by some function $f (x)$ .
How do you recognise that an equation is a hidden quadratic?
One term is the square of another.
Wherever both $f (x)$ and ${[f (x)]}^{2}$ appear alongside a constant, the equation is a quadratic in $f (x)$ .
True or False?
$\sin^{2} x + 2 \sin x - 3 = 0$ is a quadratic equation.
True.
It is a quadratic in $\sin x$ , with exactly the structure of $x^{2} + 2 x - 3 = 0$ and $x$ replaced throughout by $\sin x$ .
When you solve a hidden quadratic by substitution, why are the roots of the resulting quadratic not the final answer?
Because they are values of $f (x)$ , not values of $x$ .
Each root still has to be substituted back, so that $f (x)$ can be solved for the values of $x$ the question asked for.
When solving a hidden quadratic, what is the substitution $y = f (x)$ for?
When solving a hidden quadratic, the substitution $y = f (x)$ only makes the quadratic structure easier to see, and is never compulsory.
You can factorise directly in terms of the function instead, as in $(\cos x + 3) (\cos x - 1) = 0$ .

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