Flashcards
WJEC Core 1 Maths - Key Facts
Flashcards do conjunto (43)
- 1Formula for the gradient of a line joining two points\[ m=\frac{y_2-y_1}{x_2-x_1}\]
- 2The quadratic equation formula for solving \[ax^2+bx+c=0\]\[x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\]
- 3The 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 \( (x_2, y_2) \) - 4A line has gradient \(m\). A line perpendicular to this will have a gradient of...\[ \frac{-1}{m}\]
- 5If we know the gradient of a line and a point on the line, a formula to work out the equation of the line is...\[ y-y_1=m(x-x_1)\]
- 6Formula for the distance between two points...\[ \sqrt{\left ( x_2-x_1 \right )^2 + \left ( y_2-y_1 \right )^2} \]
- 7To find where two graphs intersect each other...... solve their equations simultaneously.
- 8In a right-angled triangle, \[ \cos{\theta}=... \]\[\frac{adjacent}{hypotenuse}\]
- 9In a right-angled triangle, \[ \sin{\theta}=... \]\[\frac{opposite}{hypotenuse}\]
- 10In a right-angled triangle, \[ \tan{\theta}=... \]\[\frac{opposite}{adjacent}\]
- 11To simplify \( \frac{a}{\sqrt{b}} \)... (a.k.a. 'rationalising the denominator')Multiply by \[ \frac{\sqrt{b}}{\sqrt{b}} \]
- 12To simplify \( \frac{a}{b+\sqrt{c}} \)... (a.k.a. 'rationalising the denominator')Multiply by \[ \frac{b-\sqrt{c}}{b-\sqrt{c}} \]
- 13\[\left(\sqrt{m} \right)^{3}=... \]\[\left(\sqrt{m} \right)^{3}=\sqrt{m}\sqrt{m}\sqrt{m}=m\sqrt{m}\]
- 14\[\sqrt{a}\times \sqrt{b}=...\]\[\sqrt{a}\times \sqrt{b}=\sqrt{ab}\]
- 15\[\frac{\sqrt{a}}{\sqrt{b}}=...\]\[\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}\]
- 16To find the gradient of a curve at any point, use...Differentiation
- 17Parallel lines have the same...Gradient
- 18To find the gradient of the line \(ax+by+c=0\)...Rearrange into the form \(y=mx+c\). The value of \(m\) is the gradient.
- 19Where is the vertex of the graph \[y=\left ( x+a \right )^2+b\]?\[\left ( -a,b \right )\]
- 20The discriminant of \(ax^2+bx+c\) is...\[b^2-4ac\]
- 21The 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 - 22If a quadratic equation has two distinct real roots, what do we know about the discriminant?\[b^2-4ac>0\]
- 23If a quadratic equation has two equal roots, what do we know about the discriminant?\[b^2-4ac=0\]
- 24If a quadratic equation has no real roots, what do we know about the discriminant?\[b^2-4ac<0\]
- 25Here 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! - 26The formula for differentiating by first principles...\[\frac{dy}{dx} = \lim_{\delta x\rightarrow 0}\left (\frac{f\left(x+\delta x\right)-f(x)}{\delta x} \right )\]
- 27If \(y=ax^n\), then \(\frac{dy}{dx} =...\)\[\frac{dy}{dx} =anx^{n-1}\]
- 28If \(\left (x+a \right )\) is a factor of \(f(x)\), then...\[f(-a)=0\]
This is known as the Factor Theorem - 29If the remainder, when \(f(x)\) is divided by \((x+a)\) is R, then...\[f(-a)=R\]
This is known as the Remainder Theorem - 30What 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 - 31What 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 - 32What 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 \(a\) - 33What 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] - 34How 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. - 35A function is said to be 'increasing' when its gradient is...Positive
- 36A function is said to be 'decreasing' when its gradient is...Negative
- 37\[a^0=?\]\[a^0=1\]
- 38\[a^m \times a^n = ?\]\[a^m \times a^n = a^{m+n}\]
- 39\[a^m \div a^n = ?\]\[a^m \div a^n = a^{m-n}\]
- 40\[\left( a^m \right) ^n=?\]\[\left( a^m \right) ^n=a^{mn}\]
- 41\[a^{-n}=?\]\[a^{-n}=\frac{1}{a^n}\]
- 42\[a^{\frac{m}{n}}=?\]\[a^{\frac{m}{n}}=\left (\sqrt[n]{a} \right )^m\]
- 43What does \(n!\) mean?\[n!=n(n-1)(n-2)\times \ldots \times 3 \times 2 \times 1\]
For example, \(4!=4\times 3\times 2\times 1=24\)
Informações
- Descrição
- Key points and important formulae for the WJEC GCE Maths C1 module
- Criado por
- Daniel Cox
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