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.

1. Start with the equation to solve: $x^2 - 2x - 2 = 0$
2. Rearrange it to make one side look like the plotted graph ($x^2 - x - 4$): $x^2 - x - 4 - x + 2 = 0$ $x^2 - x - 4 = x - 2$
3. This means we must draw the line $y = x - 2$ on the grid.
4. The intersection points of $y = x^2 - x - 4$ and $y = x - 2$ are at $x \approx -0.7$ and $x \approx 2.7$.

To use an existing graph to solve a different equation, we manipulate the target equation so that the LHS is identical to the function already drawn. The resulting RHS is the equation of the line that needs to be plotted.

1. Plot the reciprocal graph $y = \frac{4}{x}$ for positive values of $x$.
2. Plot the linear graph $y = x + 1$.
3. Locate the point of intersection in the first quadrant.
4. The graphs intersect at $(1.56, 2.56)$ approximately. Therefore, $x \approx 1.6$.

Simultaneous equations are solved graphically by finding the coordinates where the two graphs meet. For $x > 0$, we only look at the intersection in the first quadrant.