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[h] IB Mathematics AA HL Flashcards- The quadratic function

[q] **The quadratic function**

Now we have a section of the syllabus where we see some of the common types of functions. We already saw functions with just \(x^1\) and constant terms – linear functions (2.1). Now we look at functions with constants, \(x^1\) terms, and \(x^2\) terms – quadratic functions, polynomials of degree: two.

\[

f(x) = ax^2 + bx + c \quad \text{FORM}

\]

[q]

➔ Possibly the form of quadratics that you will see most often is \(f(x) = ax^2 + bx + c\), such as \(f(x) = 2x^2 – 5x + 1\) or \(y = 13 – 5x^2\). More details will be explained about this form in 2.7 – its zeros/roots. For now, we can find:

– y-intercept ➔ When \(x = 0\): \(f(0) = a(0)^2 + b(0) + c = c\)

\(\Rightarrow\) y-int. at (0, c)

[a]

– Axis of Symmetry ➔ The axis of symm. cuts through the vertex, so finding the x-coordinate of the vertex gives you the equation of the axis of symmetry. However, to show where the formula comes from, it may need calculus: \(f'(x) = 2ax + b\), set it equal to 0:

[q]

\(2ax + b = 0 \Rightarrow 2ax = -b \Rightarrow\) Axis of Symm. at \(x = \frac{-b}{2a}\)

\[

f(x) = a(x – p)(x – q) \quad \text{FORM}

\]

[a]

➔ In some ways, this form is a middle step in finding the roots of \(f(x) = ax^2 + bx + c\) through factoring. This is because it is quite easy to find the roots from this position: You may notice that to make \(f(x) = 0\), you need \(x – p = 0\) or \(x – q = 0\). This gives you \(x = p\) & \(x = q\):

[q]

➔ Roots/zeros/x-ints. at \(x = p\) & \(x = q\)

\[

f(x) = a(x – h)^2 + k \quad \text{FORM}

\]

➔ This form can be found by completing the square. \(h\) is found by doing half of \(-b\), and \(k\) is used as a correction to make the constants equal. The main benefit here is that you can find the vertex easily: When \(a > 0\), to minimise \(f(x)\), as you just need to minimise the \((x – h)^2\) part, and because squaring things always makes it positive, you must make \(x – h = 0 \Rightarrow x = h\). When \(x = h\), \(a(x – h)^2 = 0\), so \(f(x)\) has y-coordinate of \(k\):

[a]

➔ Vertex is at (h, k)

E.G. 1 ➔ If \(f(x) = 2x^2 – 4x + 7\), find the y-intercept & the axis of symmetry:

➔ y-int. at \((0, c) \Rightarrow (0, 7)\), Axis of symm. at \(x = \frac{-b}{2a} \Rightarrow x = \frac{-(-4)}{2(2)} \Rightarrow x = 1\)

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