Given that $x^2 - 6x + 1 = (x - a)^2 - b$ for all values of $x$, (i) find the value of $a$ and the value of $b$. $a = $ $b = $ (2) (ii) Hence write down the coordinates of the turning point on the graph of $y = x^2 - 6x + 1$ (1)

GCSE Edexcel Maths Direct & Inverse Proportion: Given that $x^2

Given that $x^2 - 6x + 1 = (x - a)^2 - b$ for all values of $x$, (i) find the value of $a$ and the value of $b$ - Edexcel - GCSE Maths - Question 20 - 2019 - Paper 1

Question 20

Given that $x^2 - 6x + 1 = (x - a)^2 - b$ for all values of $x$, (i) find the value of $a$ and the value of $b$. $a = $ $b = $ (2) (ii) Hence w... show full transcript

Worked Solution & Example Answer:Given that $x^2 - 6x + 1 = (x - a)^2 - b$ for all values of $x$, (i) find the value of $a$ and the value of $b$ - Edexcel - GCSE Maths - Question 20 - 2019 - Paper 1

Step 1

find the value of $a$ and the value of $b$

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Answer

To solve for $a$ and $b$ , we start by expanding the right-hand side of the equation:

$(x - a)^2 - b = x^2 - 2ax + a^2 - b.$

Next, we set the equations equal to each other:

$x^2 - 6x + 1 = x^2 - 2ax + (a^2 - b).$

By comparing coefficients, we can derive the following equations:

The coefficient of $x$ :
$-6 = -2a \Rightarrow a = 3.$
The constant term:
$1 = a^2 - b \Rightarrow 1 = 3^2 - b \Rightarrow 1 = 9 - b \Rightarrow b = 8.$

Thus, the values are:

$a = 3$ and $b = 8$ .

Step 2

Hence write down the coordinates of the turning point on the graph of $y = x^2 - 6x + 1$

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Answer

To find the coordinates of the turning point, we can use the vertex formula for a quadratic equation of the form $y = ax^2 + bx + c$ . The $x$ -coordinate of the vertex is given by:

$x = -\frac{b}{2a}.$

In our case, $a = 1$ and $b = -6$ , so:

$x = -\frac{-6}{2(1)} = 3.$

We then substitute $x = 3$ back into the original equation to find the $y$ -coordinate:

$y = (3)^2 - 6(3) + 1 = 9 - 18 + 1 = -8.$

Therefore, the coordinates of the turning point are:

$(3, -8)$ .

Given that $x^2 - 6x + 1 = (x - a)^2 - b$ for all values of $x$, (i) find the value of $a$ and the value of $b$ - Edexcel - GCSE Maths - Question 20 - 2019 - Paper 1

Worked Solution & Example Answer:Given that $x^2 - 6x + 1 = (x - a)^2 - b$ for all values of $x$, (i) find the value of $a$ and the value of $b$ - Edexcel - GCSE Maths - Question 20 - 2019 - Paper 1

find the value of $a$ and the value of $b$

Hence write down the coordinates of the turning point on the graph of $y = x^2 - 6x + 1$

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