A-Level Edexcel Maths Pure Length of an Arc: 4x^2 + 8x + 3 ≡ a(x + b)^2 + c

4x^2 + 8x + 3 ≡ a(x + b)^2 + c (a) Find the values of the constants a, b and c - Edexcel - A-Level Maths Pure - Question 11 - 2013 - Paper 3

Question 11

4x^2 + 8x + 3 ≡ a(x + b)^2 + c (a) Find the values of the constants a, b and c. (b) On the axes on page 27, sketch the curve with equation y = 4x^2 + 8x + 3, showi... show full transcript

Worked Solution & Example Answer:4x^2 + 8x + 3 ≡ a(x + b)^2 + c (a) Find the values of the constants a, b and c - Edexcel - A-Level Maths Pure - Question 11 - 2013 - Paper 3

Step 1

Find the values of the constants a, b and c.

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Answer

To find the constants $a$ , $b$ , and $c$ , we will expand the expression on the right side and compare the coefficients with the left side:

Starting from the equation:

$4x^2 + 8x + 3 ≡ a(x + b)^2 + c$
Expanding the right side:

$a(x + b)^2 = a(x^2 + 2bx + b^2) = ax^2 + 2abx + ab^2$
Thus, we have:

$ax^2 + 2abx + (ab^2 + c)$
Comparing coefficients:
- For $x^2$ terms: $a = 4$
- For $x$ terms: $2ab = 8$
- For the constant terms: $ab^2 + c = 3$
From $a = 4$ , we can substitute it into $2ab = 8$ :

$2(4)(b) = 8 \\ herefore b = 1$
Now substitute $a = 4$ and $b = 1$ into the constant equation:

$4(1)^2 + c = 3 \\ herefore 4 + c = 3 \\ herefore c = 3 - 4 = -1$

Thus, the values are:

$a = 4$
$b = 1$
$c = -1$

Step 2

On the axes on page 27, sketch the curve with equation y = 4x^2 + 8x + 3, showing clearly the coordinates of any points where the curve crosses the coordinate axes.

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Answer

To sketch the curve for the equation $y = 4x^2 + 8x + 3$ , we follow these steps:

Identify the vertex: The curve is a U-shaped graph since $a > 0$ . We can determine the vertex using the vertex formula for a quadratic equation $y = ax^2 + bx + c$ , where the x-coordinate of the vertex is given by:

$x = -\frac{b}{2a} = -\frac{8}{2(4)} = -1$

Substituting $x = -1$ back into the equation to find the y-coordinate:

$y = 4(-1)^2 + 8(-1) + 3 = 4 - 8 + 3 = -1$

Thus, the vertex is at $(-1, -1)$ .
Find x-intercepts (where $y = 0$ ):

Solving $4x^2 + 8x + 3 = 0$ using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-8 \pm \sqrt{8^2 - 4(4)(3)}}{2(4)} = \frac{-8 \pm \sqrt{64 - 48}}{8} = \frac{-8 \pm \sqrt{16}}{8} = \frac{-8 \pm 4}{8}$

Therefore, the x-intercepts are:
- $x = -\frac{1}{2}$
- $x = -\frac{3}{2}$
Find y-intercept (where $x = 0$ ):

Substituting $x = 0$ into the equation:

$y = 4(0)^2 + 8(0) + 3 = 3$

Thus, the y-intercept is at $(0, 3)$ .
Sketch the graph: Draw a U-shaped curve passing through the points:
- Vertex: $(-1, -1)$
- X-intercepts: $(-1/2, 0)$ and $(-3/2, 0)$
- Y-intercept: $(0, 3)$ .

4x^2 + 8x + 3 ≡ a(x + b)^2 + c (a) Find the values of the constants a, b and c - Edexcel - A-Level Maths Pure - Question 11 - 2013 - Paper 3

Worked Solution & Example Answer:4x^2 + 8x + 3 ≡ a(x + b)^2 + c (a) Find the values of the constants a, b and c - Edexcel - A-Level Maths Pure - Question 11 - 2013 - Paper 3

Find the values of the constants a, b and c.

On the axes on page 27, sketch the curve with equation y = 4x^2 + 8x + 3, showing clearly the coordinates of any points where the curve crosses the coordinate axes.

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