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Maths
GCSE

Sketch the curve y=4-(x+3)^2, showing the points where the curve crosses the x-axis and any minimum or maximum points.

This equation rearranges to give -y=(x+3)^2-4, which is very similar to our curve y=(x+3)^2-4 from before. In fact, replacing y with -y in an equation is equivalent to reflecting the curve through the x-a...

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Answered by Jonny I. Maths tutor
3495 Views

Find the coordinates of the minimum point of the curve y=x^2+6x+5.

To answer this question is equivalent to minimising y=(x+3)^2-4. We have that all square numbers are greater than or equal to 0 so to minimise this equation, we require that (x+3)^2=0. This is satisfied o...

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Answered by Jonny I. Maths tutor
11446 Views

Factorise x^2+6x+5=0 by completing the square.

When completing the square, we first divide the whole equation by the x^2 component. In this case, the x^2 component is 1 so nothing changes. We now apply the method to convert to square form: we reduce t...

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Answered by Jonny I. Maths tutor
8790 Views

Find the coordinates where the curve y=x^2+6x+5 crosses the x-axis.

When any curve crosses the x-axis, the y-coordinate is 0 at that point. Hence, our answers will have y=0. So we want to solve x^2+6x+5=0. From before, we have that x^2+6x+5=0 can be rewritten as (x+5)(x+1...

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Answered by Jonny I. Maths tutor
4135 Views

Factorise fully x^2+6x+5=0

To start, notice that we have x^2 in our equation. This means that our answer is of the form (x )(x ). We now look for integers that give 5 when multiplied together. The only way we can do this is with th...

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Answered by Jonny I. Maths tutor
4823 Views

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