Given the quadratic function : f(x)=x^24x-12 Find the vertex. Can you show the steps?!?

1 answer:

6 0

$\bf \textit{vertex of a vertical parabola, using coefficients}
\\\\
\begin{array}{llcccl}
\stackrel{f(x)}{y} = & 1x^2& +4x& -12\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}
\qquad 
\left(-\cfrac{ b}{2 a}\quad ,\quad c-\cfrac{ b^2}{4 a}\right)
\\\\\\
\left(-\cfrac{4}{2(1)}~~,~~-12-\cfrac{4^2}{4(1)} \right)\implies (-2~~,~~-12-4)\implies (-2~~,~~-16)$

You might be interested in

What is the area of a 4.2 yard square?

SOVA2 [1]

Answer:

37.8 square feet

Step-by-step explanation:

4 0

3 years ago

10) A wheelchair ramp is getting built at the entrance of a building. The ramp will be 25 feet long from front

mezya [45]

Answer:

The angle complies with the state regulations.

Step-by-step explanation:

The angle of elevation is the angle $\theta$ in the figure, and from trigonometry we find that

$sin(\theta) = \dfrac{1.5ft}{25ft}$

taking the inverse sine of both sides we get

$\theta = sin^{-1}(\dfrac{1.5ft}{25ft})$

which gives

$\theta = 3.44^o$

which is less than 5°, and therefore is well within the state regulations.

3 0

3 years ago

Use implicit differentiation to find the points where the parabola defined by x2−2xy+y2+4x−8y+20=0 has horizontal and vertical t

Komok [63]

Answer:

The parabola has a horizontal tangent line at the point (2,4)

The parabola has a vertical tangent line at the point (1,5)

Step-by-step explanation:

Ir order to perform the implicit differentiation, you have to differentiate with respect to x. Then, you have to use the conditions for horizontal and vertical tangent lines.

-To obtain horizontal tangent lines, the condition is:

$\frac{dy}{dx}=0$ (The slope is zero)

--To obtain vertical tangent lines, the condition is:

$\frac{dy}{dx}=\frac{1}{0}$ (The slope is undefined, therefore the denominator is set to zero)

Derivating respect to x:

$\frac{d(x^{2}-2xy+y^{2}+4x-8y+20)}{dx} = \frac{d(x^{2})}{dx}-2\frac{d(xy)}{dx}+\frac{d(y^{2})}{dx}+4\frac{dx}{dx}-8\frac{dy}{dx}+\frac{d(20)}{dx}$ = $2x -2(y+x\frac{dy}{dx})+2y\frac{dy}{dx}+4-8\frac{dy}{dx}= 0$