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Algebra - Quadratic functions in standard, vertex, and intercept forms

Grade 10IB

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

🔑Concepts

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The standard form of a quadratic function is f(x)=ax2+bx+cf(x) = ax^2 + bx + c. The coefficient aa determines if the parabola opens upwards (a>0a > 0) or downwards (a<0a < 0). The yy-intercept is always at (0,c)(0, c).

Graph of a parabola showing the y-intercept at (0, c) in standard form.
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The vertex form f(x)=a(x−h)2+kf(x) = a(x - h)^2 + k directly identifies the vertex (h,k)(h, k) and the axis of symmetry x=hx = h. This form is obtained by completing the square from the standard form.

Graph of a parabola with vertex at (h, k) and vertical axis of symmetry x = h.
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The intercept (factorized) form f(x)=a(x−p)(x−q)f(x) = a(x - p)(x - q) shows the xx-intercepts at x=px = p and x=qx = q. The axis of symmetry is located exactly halfway between the intercepts at x=p+q2x = \frac{p+q}{2}.

Parabola crossing the x-axis at points p and q.
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Transforming between forms involves algebraic manipulation. Expand vertex or intercept forms to reach standard form; use factoring to reach intercept form; use the vertex formula or completing the square to reach vertex form.

📐Formulae

f(x)=ax2+bx+cf(x) = ax^2 + bx + c

f(x)=a(x−h)2+kf(x) = a(x - h)^2 + k

f(x)=a(x−p)(x−q)f(x) = a(x - p)(x - q), where pp and qq are roots.

x=−b2ax = -\frac{b}{2a} (Axis of Symmetry)

Δ=b2−4ac\Delta = b^2 - 4ac (Discriminant to determine the number of xx-intercepts)

💡Examples

Problem 1:

Convert the quadratic function f(x)=2x2−12x+10f(x) = 2x^2 - 12x + 10 into vertex form.

Solution:

  1. Identify a=2,b=−12,c=10a = 2, b = -12, c = 10.
  2. Calculate hh: h=−−122(2)=124=3h = -\frac{-12}{2(2)} = \frac{12}{4} = 3.
  3. Calculate kk: k=f(3)=2(3)2−12(3)+10=2(9)−36+10=18−36+10=−8k = f(3) = 2(3)^2 - 12(3) + 10 = 2(9) - 36 + 10 = 18 - 36 + 10 = -8.
  4. Write in vertex form f(x)=a(x−h)2+kf(x) = a(x - h)^2 + k: f(x)=2(x−3)2−8f(x) = 2(x - 3)^2 - 8.

Explanation:

To convert from standard to vertex form, we find the coordinates of the vertex (h,k)(h, k) using the formula x=−b2ax = -\frac{b}{2a} and substituting that value back into the original function.

Problem 2:

Find the equation of a parabola in intercept form that has xx-intercepts at x=−2x = -2 and x=4x = 4 and passes through the point (0,−16)(0, -16).

Solution:

  1. Start with the intercept form: f(x)=a(x−p)(x−q)f(x) = a(x - p)(x - q).
  2. Substitute the intercepts: f(x)=a(x−(−2))(x−4)=a(x+2)(x−4)f(x) = a(x - (-2))(x - 4) = a(x + 2)(x - 4).
  3. Use the point (0,−16)(0, -16) to find aa: −16=a(0+2)(0−4)-16 = a(0 + 2)(0 - 4).
  4. Solve for aa: −16=a(2)(−4)⇒−16=−8a⇒a=2-16 = a(2)(-4) \Rightarrow -16 = -8a \Rightarrow a = 2.
  5. The final equation is f(x)=2(x+2)(x−4)f(x) = 2(x + 2)(x - 4).

Explanation:

By substituting the known roots into the intercept form, we create a template. We then use the provided (x,y)(x, y) coordinate to solve for the stretch factor aa.

Problem 3:

Identify the features of the function f(x)=−(x+1)2+9f(x) = -(x + 1)^2 + 9: Vertex, Axis of Symmetry, and direction of opening.

Solution:

  1. Compare to f(x)=a(x−h)2+kf(x) = a(x - h)^2 + k: here a=−1,h=−1,k=9a = -1, h = -1, k = 9.
  2. Vertex: (−1,9)(-1, 9).
  3. Axis of Symmetry: x=−1x = -1.
  4. Direction: Since a=−1a = -1 (which is <0< 0), the parabola opens downwards.

Explanation:

In vertex form, hh is the value subtracted from xx, so (x+1)(x+1) implies h=−1h = -1. The value of kk is added at the end. The sign of aa indicates whether the parabola is a 'U' shape or an 'n' shape.

Problem 4:

Determine the equation of the quadratic function shown in the graph, which has a vertex at (2,−4)(2, -4) and passes through the point (0,0)(0, 0). Express the answer in vertex form.

A parabola with vertex at (2, -4) passing through the origin.

Solution:

  1. Start with the vertex form: f(x)=a(x−h)2+kf(x) = a(x - h)^2 + k.
  2. Substitute the vertex (2,−4)(2, -4): f(x)=a(x−2)2−4f(x) = a(x - 2)^2 - 4.
  3. Use the point (0,0)(0, 0) to solve for aa: 0=a(0−2)2−40 = a(0 - 2)^2 - 4 0=4a−40 = 4a - 4 4=4a  ⟹  a=14 = 4a \implies a = 1.
  4. The equation is f(x)=(x−2)2−4f(x) = (x - 2)^2 - 4.

Explanation:

Vertex form is the most efficient choice when the turning point is known. Substituting another known point allows us to solve for the vertical stretch factor aa.

Problem 5:

A quadratic function is given by f(x)=−x2+4x+5f(x) = -x^2 + 4x + 5. Find its xx-intercepts by converting it to intercept form and sketch the graph.

Downward opening parabola with intercepts at -1 and 5.

Solution:

  1. Factor out the negative sign: f(x)=−(x2−4x−5)f(x) = -(x^2 - 4x - 5).
  2. Factor the quadratic expression: f(x)=−(x−5)(x+1)f(x) = -(x - 5)(x + 1).
  3. Set f(x)=0f(x) = 0 to find intercepts: x−5=0x - 5 = 0 or x+1=0x + 1 = 0.
  4. The xx-intercepts are (5,0)(5, 0) and (−1,0)(-1, 0).

Explanation:

Factoring the standard form into intercept form reveals the roots of the function. Since a=−1a = -1, the parabola opens downwards.