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3 years ago

Explain how the value of a in the standard form of a quadratic affects the direction and shape of the graph.

Mathematics

1 answer:

julsineya [31]3 years ago

6 0

Answer:

Affects the steepness of the graph

Step-by-step explanation:

A quadratic is in the form ax^2 + bx + c and is usually a parabola.

Changing the constant a will influence the gradient of this parabola and will therefore affect how steep it is.

If a is positive, the parabola will have a minimum turning point meaning that it will be facing upwards.

if a is negative, it will have a maximum turning point and face downwards.

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baherus [9]

Answer: see proof below

<u>Step-by-step explanation:</u>

Given: cos 330 = $\frac{\sqrt3}{2}$

Use the Double-Angle Identity: cos 2A = 2 cos² A - 1

$\text{Scratchwork:}\quad \bigg(\dfrac{\sqrt3 + 2}{2\sqrt2}\bigg)^2 = \dfrac{2\sqrt3 + 4}{8}$

Proof LHS → RHS:

LHS cos 165

Double-Angle: cos (2 · 165) = 2 cos² 165 - 1

⇒ cos 330 = 2 cos² 165 - 1

⇒ 2 cos² 165 = cos 330 + 1

Given: $2 \cos^2 165 = \dfrac{\sqrt3}{2} + 1$

$\rightarrow 2 \cos^2 165 = \dfrac{\sqrt3}{2} + \dfrac{2}{2}$

Divide by 2: $\cos^2 165 = \dfrac{\sqrt3+2}{4}$

$\rightarrow \cos^2 165 = \bigg(\dfrac{2}{2}\bigg)\dfrac{\sqrt3+2}{4}$

$\rightarrow \cos^2 165 = \dfrac{2\sqrt3+4}{8}$

Square root: $\sqrt{\cos^2 165} = \sqrt{\dfrac{4+2\sqrt3}{8}}$

Scratchwork: $\cos^2 165 = \bigg(\dfrac{\sqrt3+1}{2\sqrt2}\bigg)^2$

$\rightarrow \cos 165 = \pm \dfrac{\sqrt3+1}{2\sqrt2}$

Since cos 165 is in the 2nd Quadrant, the sign is NEGATIVE

$\rightarrow \cos 165 = - \dfrac{\sqrt3+1}{2\sqrt2}$

LHS = RHS $\checkmark$

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