Given the expression $y = x^2 + 2x + 3 = (x + a)^2 + b$ - Edexcel - A-Level Maths Pure - Question 5 - 2006 - Paper 1

The graph is a 'U'-shaped parabola with the vertex located at the point $(0, 3)$.

To find intersections with the axes:

1. Y-intercept: Set $x = 0$, then $y = 3$.
2. X-intercept: There are no x-intercepts since the discriminant is negative (as found in part c).

The graph will lie entirely above the x-axis due to this negative discriminant.

The discriminant of a quadratic $ax^2 + bx + c$ is given by the formula: $D = b^2 - 4ac$

For $x^2 + 2x + 3$, we have $a = 1$, $b = 2$, and $c = 3$: $D = 2^2 - 4 imes 1 imes 3 = 4 - 12 = -8$

The negative discriminant indicates that there are no real roots, which aligns with our graph, confirming that the curve does not intersect the x-axis.

To find values of $k$ for the equation $x^2 + kx + 3 = 0$ to have no real roots, we again use the discriminant condition:

1. The discriminant must be negative: $D = k^2 - 4 imes 1 imes 3 < 0$ This simplifies to: $k^2 - 12 < 0$ $k^2 < 12$ Taking square roots gives: $$$$

t{12} < k < t{12}$$ Therefore, the possible values of $k$ are: