Algebra and Graphs - Solving equations by graphical methods

The graph of $y = x^2 - 4x + 2$ is already drawn. Use the graph to solve the equation $x^2 - 4x - 1 = 0$.

1. Rearrange the target equation to match the graph: $x^2 - 4x + 2 - 3 = 0$.
2. This becomes $x^2 - 4x + 2 = 3$.
3. Draw the horizontal line $y = 3$ on the same axes as the curve.
4. Identify the x-coordinates where the line $y = 3$ intersects the curve $y = x^2 - 4x + 2$.
5. Solutions are $x \approx -0.2$ and $x \approx 4.2$.

To use an existing graph $y = f(x)$ to solve a new equation, we must manipulate the new equation until one side is identical to $f(x)$. The other side of the equation tells us what additional line or curve needs to be drawn.

Estimate the gradient of the curve $y = x^3 - 3x$ at the point where $x = 2$.

1. Locate the point $(2, 2)$ on the graph of $y = x^3 - 3x$.
2. Use a ruler to draw a tangent line that just touches the curve at this point.
3. Pick two points on this tangent line, e.g., $(1, -7)$ and $(3, 11)$.
4. Calculate $m = \frac{11 - (-7)}{3 - 1} = \frac{18}{2} = 9$.
5. The estimated gradient is 9.

The gradient of a curve changes at every point. A tangent line represents the instantaneous rate of change at that specific point. Accuracy depends on the precision of the drawn tangent.

Find the range of values of $k$ for which the equation $x^2 - 4x + 2 = k$ has no real solutions.

1. Observe the vertex (minimum point) of the parabola $y = x^2 - 4x + 2$.
2. The vertex is at $x = -b/2a = 4/2 = 2$.
3. $y$-value at vertex: $(2)^2 - 4(2) + 2 = -2$.
4. If the line $y = k$ is below the minimum point, there are no intersections.
5. Therefore, $k < -2$.

Graphically, 'no real solutions' means the line $y = k$ does not intersect the curve $y = f(x)$. For a U-shaped quadratic, this occurs when $k$ is less than the minimum $y$-value.