f(x) = x² + 4kx + (3 + 11k), where k is a constant - Edexcel - A-Level Maths Pure - Question 2 - 2009 - Paper 1

The condition for the quadratic equation f(x) = 0 to have no real roots is that the discriminant must be less than zero.

The discriminant (D) is given by: $D = b² - 4ac$ For our function, ( a = 1, b = 4k, c = (3 + 11k - 4k²) ) Thus, $D = (4k)² - 4(1)(3 + 11k - 4k²)$

Setting up the inequality: $16k² - 4(3 + 11k - 4k²) < 0$

Expanding and simplifying gives:

• Rearranging tells us that: $4k² - 11k + 3 < 0$
• Factoring reveals: $(k - 3)(k - \frac{1}{4}) < 0$

The solutions to this inequality reveal that:

• The critical points are ( k = \frac{1}{4} \text{ and } 3 )

Thus, the set of possible values of k is: $\frac{1}{4} < k < 3$

Setting k = 1, we have:

$f(x) = x² + 4(1)x + (3 + 11(1))$ $f(x) = x² + 4x + 14$

To find points where the graph crosses the axes, we:

1. Find x-intercepts by setting ( f(x) = 0 ) and solving for x:

$x² + 4x + 14 = 0$

The discriminant is: $D = 4² - 4(1)(14) = 16 - 56 = -40$

This means there are no x-intercepts (the curve does not cross the x-axis).

2. Find the y-intercept by evaluating f(0):

$f(0) = 14$

Thus, the graph crosses the y-axis at (0, 14).

Graph Characteristics:

• The parabola opens upwards.
• It touches the y-axis at (0, 14).
• The vertex can be found at ( x = -\frac{b}{2a} = -\frac{4}{2} = -2 ). Evaluating f(-2) gives the minimum value.

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