P1.1 - Algebra and functions

Index laws describe repeated multiplication and remain consistent for zero, negative and rational exponents. Before combining powers, check that they have the same base.

To solve $3^{2p-5}=3^{-1/2}$, equal positive bases give equal exponents: $2p-5=-\tfrac12$, so $p=\tfrac94$.

Rewrite composite bases such as 9 or 27 as powers of one base, simplify each side fully, then equate exponents only after the bases match.

A surd is an exact irrational root. Simplify it by extracting perfect-square factors, combine only like surds, and rationalise a denominator without changing the value.

Since $(2+\sqrt3)(2-\sqrt3)=1$, $\dfrac1{2+\sqrt3}=2-\sqrt3$. The conjugate removes the surd from the denominator while multiplying by 1.

For example, $3\sqrt8-\sqrt{18}=3(2\sqrt2)-3\sqrt2=3\sqrt2$. Simplify each root before deciding whether terms are like surds.

A quadratic function $f(x)=ax^2+bx+c$, $a\ne0$, has a parabolic graph. Its algebraic form can expose intercepts, symmetry or the turning point before any points are plotted.

For $y=(x-1)(x-5)$, the roots are 1 and 5, so the axis is $x=3$. Substitution gives $y(3)=-4$, hence the minimum is $(3,-4)$; the positive leading coefficient means the graph opens upward.

Mark the turning point, intercepts and symmetry axis, then join them with one smooth parabola. If there are no real roots, the graph remains entirely above or below the horizontal axis according to its turning point and opening.

For $x^2-4x+k=0$ to have no real roots, $\Delta=16-4k<0$, so $k>4$. The inequality on the discriminant becomes the condition on the parameter.

First rearrange the equation into quadratic form and identify $a$, $b$ and $c$ with their signs. For tangency use equality; for ‘at least one real root’ use $\Delta\ge0$.

Factor out the leading coefficient from the $x$ terms, take half the coefficient of $x$ inside the bracket, square it, and compensate for the added term. The resulting form gives the turning point and supports exact solution.

$2x^2-8x+3=2(x^2-4x)+3=2[(x-2)^2-4]+3=2(x-2)^2-5$. Hence $2x^2-8x+3=0$ gives $(x-2)^2=5/2$ and $x=2\pm\sqrt{10}/2$.

Use factorisation when factors are visible, completing the square when vertex form or an exact rearrangement is valuable, and the quadratic formula as a general exact method. A calculator can confirm but not replace required algebraic stages.

A simultaneous solution must satisfy both equations. Isolate one variable, substitute its complete expression into the other equation, solve for every possible value, then back-substitute to form ordered pairs.

For $y=x+1$ and $x^2+y^2=13$, substitution gives $x^2+(x+1)^2=13$, so $x^2+x-6=0$. Thus $x=2$ or $-3$, producing $(2,3)$ and $(-3,-2)$.

A line and a quadratic curve may meet twice, once or not at all. The number of real substituted roots matches the number of real intersection points.

To interpret $f(x)<g(x)$ graphically, find where the graphs meet and read the $x$-intervals on which the graph of $f$ lies below the graph of $g$. Equality determines whether intersection endpoints are included.

$x^2<x+2$ compares the parabola $y=x^2$ with the line $y=x+2$. They intersect at $x=-1$ and $x=2$; the parabola lies below the line between them, so $-1<x<2$.

Retain domain restrictions when reducing a fractional inequality. For example, multiplying $a/x<b$ by the always-positive $x^2$ gives $ax<bx^2$ together with $x\ne0$.

A graphical inequality is a region separated by a boundary curve or line. Draw the boundary first, decide whether it belongs to the region, then use a test point to choose the correct side.

For $y\ge x^2$ and $y<x+2$, draw the parabola solid and the line dotted. Shade points on or above the parabola and below the line; the common region exists between their intersection values $x=-1$ and $x=2$.

For several inequalities, shade only the overlap satisfying every condition. A vertical boundary such as $x\le3$ is handled by the same solid/dotted and test-point rules.

Move every term to one side so the problem becomes a sign question. Find the critical roots, split the number line at those roots, and determine where the expression has the required sign.

$(x-4)(x+1)\le0$ has roots $-1$ and $4$. The upward-opening product is non-positive between the roots, including zeros, so $-1\le x\le4$.

For a quadratic compared with a line, bring the line to the same side first: $x^2<3x+4$ becomes $x^2-3x-4<0$, then factor and apply the sign method.

Expanding, collecting and factorising are equivalent rewritings of a polynomial. Preserve every term and sign so the new expression has the same value for all $x$.

$x^3+4x^2+3x=x(x^2+4x+3)=x(x+1)(x+3)$. The factorised form shows zeros $0,-1,-3$, while expansion confirms the identity.

The notation $f(x)$ names the output of a polynomial rule. Solving $f(x)=g(x)$ requires bringing all terms to one side, simplifying, then factorising the resulting polynomial of degree at most 3 in this unit.

A function sketch records intercepts, symmetry, turning behaviour, periodicity and asymptotes. Intersections of $y=f(x)$ and $y=g(x)$ are the geometric solutions of $f(x)=g(x)$.

An asymptote is a line the curve approaches according to its limiting behaviour. It controls the sketch even when it is not part of the graph; reciprocal functions are undefined at $x=0$.

To find exact intersection coordinates, solve $f(x)=g(x)$ algebraically, then substitute each valid $x$ into either function for $y$. A sketch predicts how many real solutions to expect and helps detect missing roots.

A graph transformation moves each point of $y=f(x)$ by changing either its output or its input. Output changes act vertically; input changes act horizontally in the inverse-looking direction.

If $(3,5)$ lies on $y=f(x)$, then $(1,5)$ lies on $y=f(x+2)$, while $(3,10)$ lies on $y=2f(x)$. Transform intercepts, turning points and asymptotes by the same mapping.

Apply the one stated transformation to the whole curve, preserving its connections and characteristic shape. For periodic graphs, transform wavelength/period and amplitude consistently rather than moving isolated peaks.