# IB Mathematics AA HL Flashcards- The quadratic function

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

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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.

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

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➔ 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)

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– 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:

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

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

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➔ 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 $$x = p$$ & $$x = q$$:

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➔ 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 $$k$$:

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➔ 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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