krit.club logo

Algebra and Graphs - Solving equations by graphical methods

Grade 11IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Graphical Solutions: The solutions (roots) of the equation f(x)=0f(x) = 0 are the x-coordinates of the points where the graph y=f(x)y = f(x) intersects the x-axis.

•

Intersection of two functions: To solve f(x)=g(x)f(x) = g(x), find the x-coordinates of the points where the graphs of y=f(x)y = f(x) and y=g(x)y = g(x) intersect.

•

Solving f(x)=kf(x) = k: The solutions are found by identifying where the curve y=f(x)y = f(x) crosses the horizontal line y=ky = k.

•

Equation Rearrangement: If you have a graph of y=f(x)y = f(x) and need to solve a different equation like f(x)+h(x)=0f(x) + h(x) = 0, rearrange it to f(x)=−h(x)f(x) = -h(x) and draw the line y=−h(x)y = -h(x).

•

Estimation of Gradients: The gradient of a curve at a specific point is found by drawing a tangent to the curve at 그 point and calculating its slope.

•

Turning Points: Maxima and minima are the points where the gradient of the graph is zero (horizontal tangent).

📐Formulae

y=mx+cy = mx + c (Equation of the straight line to be drawn)

m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1} (Gradient calculation for a tangent line)

f(x)=g(x)  ⟹  Intersection points of y=f(x) and y=g(x)f(x) = g(x) \implies \text{Intersection points of } y=f(x) \text{ and } y=g(x)

ax2+bx+c=0ax^2 + bx + c = 0 (Standard quadratic to be solved via intercepts)

💡Examples

Problem 1:

The graph of y=x2−4x+2y = x^2 - 4x + 2 is already drawn. Use the graph to solve the equation x2−4x−1=0x^2 - 4x - 1 = 0.

Solution:

  1. Rearrange the target equation to match the graph: x2−4x+2−3=0x^2 - 4x + 2 - 3 = 0.
  2. This becomes x2−4x+2=3x^2 - 4x + 2 = 3.
  3. Draw the horizontal line y=3y = 3 on the same axes as the curve.
  4. Identify the x-coordinates where the line y=3y = 3 intersects the curve y=x2−4x+2y = x^2 - 4x + 2.
  5. Solutions are x≈−0.2x \approx -0.2 and x≈4.2x \approx 4.2.

Explanation:

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

Problem 2:

Estimate the gradient of the curve y=x3−3xy = x^3 - 3x at the point where x=2x = 2.

Solution:

  1. Locate the point (2,2)(2, 2) on the graph of y=x3−3xy = 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)(1, -7) and (3,11)(3, 11).
  4. Calculate m=11−(−7)3−1=182=9m = \frac{11 - (-7)}{3 - 1} = \frac{18}{2} = 9.
  5. The estimated gradient is 9.

Explanation:

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.

Problem 3:

Find the range of values of kk for which the equation x2−4x+2=kx^2 - 4x + 2 = k has no real solutions.

Solution:

  1. Observe the vertex (minimum point) of the parabola y=x2−4x+2y = x^2 - 4x + 2.
  2. The vertex is at x=−b/2a=4/2=2x = -b/2a = 4/2 = 2.
  3. yy-value at vertex: (2)2−4(2)+2=−2(2)^2 - 4(2) + 2 = -2.
  4. If the line y=ky = k is below the minimum point, there are no intersections.
  5. Therefore, k<−2k < -2.

Explanation:

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

Solving equations by graphical methods Revision - Grade 11 Maths IGCSE