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Algebra and Graphs - Linear and quadratic equations

Grade 11A Level

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

🔑Concepts

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The gradient (slope) of a linear graph represents the rate of change. For a line y=mx+cy = mx + c, mm is the gradient and cc is the yy-intercept where the line crosses the yy-axis.

Graph of a linear equation showing the y-intercept at (0,1) and a positive gradient.
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A quadratic graph y=ax2+bx+cy = ax^2 + bx + c forms a parabola. If a>0a > 0, the parabola opens upwards (minimum point); if a<0a < 0, it opens downwards (maximum point). The xx-intercepts are found by solving ax2+bx+c=0ax^2 + bx + c = 0.

Parabola opening upwards with vertex at (0,-4) and roots at x=2 and x=-2.
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Solving simultaneous equations graphically involves finding the intersection point(s). A linear equation and a quadratic equation can intersect at zero, one, or two points.

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The discriminant Δ=b2−4ac\Delta = b^2 - 4ac determines the nature of the roots of a quadratic equation. If Δ>0\Delta > 0, there are two distinct real roots (two xx-intercepts). If Δ=0\Delta = 0, there is one repeated real root (the graph touches the xx-axis). If Δ<0\Delta < 0, there are no real roots.

📐Formulae

y=mx+cy = mx + c (Gradient-intercept form)

m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1} (Gradient formula)

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} (The Quadratic Formula)

(x+p)2+q=0(x + p)^2 + q = 0 (Completed Square form)

y=a(x−h)2+ky = a(x - h)^2 + k (Vertex form of a quadratic graph)

💡Examples

Problem 1:

Solve the quadratic equation 2x2−7x+3=02x^2 - 7x + 3 = 0 using the quadratic formula.

Solution:

x=−(−7)±(−7)2−4(2)(3)2(2)=7±49−244=7±254x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(3)}}{2(2)} = \frac{7 \pm \sqrt{49 - 24}}{4} = \frac{7 \pm \sqrt{25}}{4}. Therefore, x=7+54=3x = \frac{7+5}{4} = 3 or x=7−54=0.5x = \frac{7-5}{4} = 0.5.

Explanation:

Identify a=2,b=−7,c=3a=2, b=-7, c=3. Substitute these into the quadratic formula. Simplify the discriminant (sqrt25=5\\sqrt{25} = 5) and solve for both the plus and minus cases.

Problem 2:

Solve the simultaneous equations: y=2x−3y = 2x - 3 and y=x2−4x+5y = x^2 - 4x + 5.

Solution:

2x−3=x2−4x+5⇒x2−6x+8=02x - 3 = x^2 - 4x + 5 \Rightarrow x^2 - 6x + 8 = 0. Factorising gives (x−2)(x−4)=0(x-2)(x-4)=0, so x=2x=2 or x=4x=4. When x=2,y=2(2)−3=1x=2, y=2(2)-3=1. When x=4,y=2(4)−3=5x=4, y=2(4)-3=5. Solutions: (2,1)(2, 1) and (4,5)(4, 5).

Explanation:

Since both equations are equal to yy, set them equal to each other to form a single quadratic equation. Solve for xx, then substitute xx back into the linear equation to find the corresponding yy values.

Problem 3:

Find the coordinates of the turning point (vertex) of the graph y=x2−6x+10y = x^2 - 6x + 10 by completing the square.

Solution:

y=(x−3)2−(−3)2+10=(x−3)2−9+10=(x−3)2+1y = (x - 3)^2 - (-3)^2 + 10 = (x - 3)^2 - 9 + 10 = (x - 3)^2 + 1. The vertex is (3,1)(3, 1).

Explanation:

Halve the coefficient of xx (which is −6-6) to get −3-3 for the bracket (x−3)2(x-3)^2. Subtract the square of that number and add the constant. In the form (x−h)2+k(x-h)^2 + k, the vertex is (h,k)(h, k).

Problem 4:

Solve the quadratic equation x2−2x−3=0x^2 - 2x - 3 = 0 by factoring and sketch the graph of y=x2−2x−3y = x^2 - 2x - 3.

Parabola for y = x^2 - 2x - 3 showing roots at -1 and 3, and vertex at (1,-4).

Solution:

  1. Factorize the quadratic: (x−3)(x+1)=0(x - 3)(x + 1) = 0
  2. Solve for xx: x−3=0  ⟹  x=3x - 3 = 0 \implies x = 3 x+1=0  ⟹  x=−1x + 1 = 0 \implies x = -1
  3. Identify the yy-intercept: When x=0,y=−3x=0, y=-3.
  4. Find the vertex: x=−b2a=−(−2)2(1)=1x = \frac{-b}{2a} = \frac{-(-2)}{2(1)} = 1. y=(1)2−2(1)−3=−4y = (1)^2 - 2(1) - 3 = -4. Vertex is (1,−4)(1, -4).

Explanation:

Factoring allows us to find the roots (x-intercepts) easily. The vertex is found using the axis of symmetry formula x=−b/2ax = -b/2a, which provides the minimum point for this upward-opening parabola.

Problem 5:

Determine the points of intersection for the linear equation y=x+2y = x + 2 and the quadratic equation y=x2−4y = x^2 - 4. Sketch the graphs to verify your solution.

Graph showing the intersection of the line y = x + 2 and the parabola y = x^2 - 4 at points (-2, 0) and (3, 5).

Solution:

  1. Set the equations equal to each other to find the xx-coordinates of the intersection points: x2−4=x+2x^2 - 4 = x + 2
  2. Rearrange into a standard quadratic equation ax2+bx+c=0ax^2 + bx + c = 0: x2−x−6=0x^2 - x - 6 = 0
  3. Factorize the quadratic: (x−3)(x+2)=0(x - 3)(x + 2) = 0
  4. Solve for xx: x=3 or x=−2x = 3 \text{ or } x = -2
  5. Substitute xx values back into the linear equation y=x+2y = x + 2 to find yy: If x=3x = 3, y=3+2=5y = 3 + 2 = 5. Point: (3,5)(3, 5) If x=−2x = -2, y=−2+2=0y = -2 + 2 = 0. Point: (−2,0)(-2, 0)
  6. The intersection points are (3,5)(3, 5) and (−2,0)(-2, 0).

Explanation:

To find where a line and a curve meet, we solve them simultaneously. By setting the expressions for yy equal, we create a single quadratic equation in terms of xx. The solutions to this quadratic provide the xx-coordinates of the intersections. Substituting these into the simpler linear equation gives the corresponding yy-coordinates.