Which function has only one x-intercept at (-6, 0)?

2 answers:

Lelu [443]3 years ago

6 0

Answer:

<h2>The function f(x) = (x - 6)(x - 6) has only one x-intercept. But at (6, 0) not at (-6, 0).</h2>

Step-by-step explanation:

The intercept form of a quadratic equation (parabola):

$y=a(x-p)(x-q)$

p, q - x-intercepts

Therefore

The function f(x) = x(x - 6) = (x - 0)(x - 6) has two x-intercepts at (0, 0) and (6, 0)

The function f(x) = (x - 6)(x - 6) has only one x-intercept at (6, 0)

The function f(x) = (x + 6)(x - 6) = (x - (-6))(x - 6)

has two x-intercept at (-6, 0) and (6, 0)

The function f(x) = (x + 1)(x + 6) = (x - (-1))(x - (-6))

has two x-intercepts at (-1, 0) and (-6, 0).

3 0

Answer:

the answer is d on edge

Step-by-step explanation:

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nikklg [1K]

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RoseWind [281]

Answer:

$\displaystyle \frac{dy}{dx} = -2x \tan (x^2)$

General Formulas and Concepts:

Calculus

Differentiation

Basic Power Rule:

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle y = \ln (\cos x^2)$

Step 2: Differentiate

Logarithmic Differentiation [Derivative Rule - Chain Rule]: $\displaystyle y' = \frac{(\cos x^2)'}{\cos x^2}$
Trigonometric Differentiation [Derivative Rule - Chain Rule]: $\displaystyle y' = \frac{-\sin x^2 (x^2)'}{\cos x^2}$
Basic Power Rule: $\displaystyle y' = \frac{-2x \sin x^2}{\cos x^2}$
Rewrite [Trigonometric Identities]: $\displaystyle y' = -2x \tan (x^2)$