WJEC Core 1 Maths - Key Facts

Description

Key points and important formulae for the WJEC GCE Maths C1 module
Daniel Cox
Flashcards by Daniel Cox, updated more than 1 year ago
Daniel Cox
Created by Daniel Cox almost 8 years ago
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Resource summary

Question Answer
Formula for the gradient of a line joining two points \[ m=\frac{y_2-y_1}{x_2-x_1}\]
The quadratic equation formula for solving \[ax^2+bx+c=0\] \[x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\]
The midpoint of \( (x_1, y_1) \) and \( (x_2, y_2) \) is... \[ \left ( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right ) \] Think of this as the mean of the coordinates \( (x_1, y_1) \) and \( (x_2, y_2) \)
A line has gradient \(m\). A line perpendicular to this will have a gradient of... \[ \frac{-1}{m}\]
If we know the gradient of a line and a point on the line, a formula to work out the equation of the line is... \[ y-y_1=m(x-x_1)\]
Formula for the distance between two points... \[ \sqrt{\left ( x_2-x_1 \right )^2 + \left ( y_2-y_1 \right )^2} \]
To find where two graphs intersect each other... ... solve their equations simultaneously.
In a right-angled triangle, \[ \cos{\theta}=... \] \[\frac{adjacent}{hypotenuse}\]
In a right-angled triangle, \[ \sin{\theta}=... \] \[\frac{opposite}{hypotenuse}\]
In a right-angled triangle, \[ \tan{\theta}=... \] \[\frac{opposite}{adjacent}\]
To simplify \( \frac{a}{\sqrt{b}} \)... (a.k.a. 'rationalising the denominator') Multiply by \[ \frac{\sqrt{b}}{\sqrt{b}} \]
To simplify \( \frac{a}{b+\sqrt{c}} \)... (a.k.a. 'rationalising the denominator') Multiply by \[ \frac{b-\sqrt{c}}{b-\sqrt{c}} \]
\[\left(\sqrt{m} \right)^{3}=... \] \[\left(\sqrt{m} \right)^{3}=\sqrt{m}\sqrt{m}\sqrt{m}=m\sqrt{m}\]
\[\sqrt{a}\times \sqrt{b}=...\] \[\sqrt{a}\times \sqrt{b}=\sqrt{ab}\]
\[\frac{\sqrt{a}}{\sqrt{b}}=...\] \[\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}\]
To find the gradient of a curve at any point, use... Differentiation
Parallel lines have the same... Gradient
To find the gradient of the line \(ax+by+c=0\)... Rearrange into the form \(y=mx+c\). The value of \(m\) is the gradient.
Where is the vertex of the graph \[y=\left ( x+a \right )^2+b\]? \[\left ( -a,b \right )\]
The discriminant of \(ax^2+bx+c\) is... \[b^2-4ac\]
The discriminant of a quadratic equation tells us... How many roots (or solutions) it has. This will be how many times it crosses the \(x\)-axis
If a quadratic equation has two distinct real roots, what do we know about the discriminant? \[b^2-4ac>0\]
If a quadratic equation has two equal roots, what do we know about the discriminant? \[b^2-4ac=0\]
If a quadratic equation has no real roots, what do we know about the discriminant? \[b^2-4ac<0\]
Here is the graph of \(y=x^2-8x+7\). Use it to solve the quadratic inequality \(x^2-8x+7>0\) \(x<1\) or \(x>7\) These are the red sections of the curve. Note - do not write \(x<1\) and \(x>7\) - the word 'and' implies \(x\) would need to be \(<1\) and \(>7\) at the same time... which is clearly not possible!
The formula for differentiating by first principles... \[\frac{dy}{dx} = \lim_{\delta x\rightarrow 0}\left (\frac{f\left(x+\delta x\right)-f(x)}{\delta x} \right )\]
If \(y=ax^n\), then \(\frac{dy}{dx} =...\) \[\frac{dy}{dx} =anx^{n-1}\]
If \(\left (x+a \right )\) is a factor of \(f(x)\), then... \[f(-a)=0\] This is known as the Factor Theorem
If the remainder, when \(f(x)\) is divided by \((x+a)\) is R, then... \[f(-a)=R\] This is known as the Remainder Theorem
What effect will the transformation \(y=f(x)+a\) have on the graph of \(y=f(x)\)? Translation \(a\) units in the \(y\) direction. i.e. the graph will move UP by \(a\) units
What effect will the transformation \(y=f(x+a)\) have on the graph of \(y=f(x)\)? Translation \(-a\) units in the \(x\) direction. i.e. the graph will move LEFT by \(a\) units
What effect will the transformation \(y=af(x)\) have on the graph of \(y=f(x)\)? Stretch, scale factor \(a\) in the \(y\) direction. i.e. the \(y\) values will be multiplied by \(a\)
What effect will the transformation \(y=f(ax)\) have on the graph of \(y=f(x)\)? Stretch, scale factor \(\frac{1}{a}\) in the \(x\) direction. i.e. the \(x\) values will be divided by \(a\) [This could also be described as a 'squash', scale factor \(a\) in the \(x\) direction]
How would you use the second derivative, \(\frac{d^2 y}{dx^2}\) to determine the nature of the stationary points on a graph? Substitute the \(x\) co-ordinates of the stationary points into \(\frac{d^2 y}{dx^2}\). If you get a positive answer, it's a MIN. If you get a negative answer, it's a MAX.
A function is said to be 'increasing' when its gradient is... Positive
A function is said to be 'decreasing' when its gradient is... Negative
\[a^0=?\] \[a^0=1\]
\[a^m \times a^n = ?\] \[a^m \times a^n = a^{m+n}\]
\[a^m \div a^n = ?\] \[a^m \div a^n = a^{m-n}\]
\[\left( a^m \right) ^n=?\] \[\left( a^m \right) ^n=a^{mn}\]
\[a^{-n}=?\] \[a^{-n}=\frac{1}{a^n}\]
\[a^{\frac{m}{n}}=?\] \[a^{\frac{m}{n}}=\left (\sqrt[n]{a} \right )^m\]
What does \(n!\) mean? \[n!=n(n-1)(n-2)\times \ldots \times 3 \times 2 \times 1\] For example, \(4!=4\times 3\times 2\times 1=24\)
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