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3 years ago

Can anyone rewrite quadratic function in standard form?

Mathematics

2 answers:

N76 [4]3 years ago

8 0

Quadratic function in standard form should be f(x)=a(x-h)^2+k

kati45 [8]3 years ago

4 0

Answer:

hey I believe standard form would look like this: f(x) = a(x - h)^2 + k

this is regular form f(x)=ax^2+bx+c or something like x^2+4x+4.

I think H and K are the vertex

hope this is close to what your looking for.

Step-by-step explanation:

I think it's x^2+6x-18

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2 years ago

PLEASE ANSWER ASAP!!

Rufina [12.5K]

Answer:

The instantaneous rate of increase of f(x) at $x_{0} = 3$ is 3.
One possible equation of this line is y = 3x - 16.
The line is tangent to the graph of y = f(x). The slope of the line is the same as the instantaneous rate of increase of f(x) at $x_{0} = 3$ .

Step-by-step explanation:

$\begin{aligned}&\lim_{h\to 0}{\frac{f(3+h)-f(h)}{h}}\\ =&\lim_{h\to 0}\frac{1}{h} \cdot [(3^{3} + 3 \times 3^{2}h +3\times 3h^{2} + h^{3})- 4(3^{2} + 2\times 3h + h^{2}) + 2 \\[-1em] &\phantom{\lim_{h\to 0}\frac{1}{h}\cdot []} -(3^{3} -4\times 3^{2} + 2)]\\ = &\lim_{h \to 0}\frac{h^{3} + 5h^{2} +3h}{h}\\=& \lim_{h \to 0}{h^{2}} + \lim_{h \to 0}{5h} + \lim_{h\to 0}{3}\\ =& 3\end{aligned}$ .

$f(3) = 3^{3} - 4\times 3^{2} + 2 =-7$ .

In other words, the graph of y = f(x) passes through the point (3, -7) where $x = 3$ .

The point-slope form of a line in a cartesian plane is:

$y - y_0 = m (x - x_0)$ .

For this line,

$(3, -7)$ is the point on the line, while

$3$ is the slope of the line.

The equation of this line will thus be

$y - (-7) = 3(x - 3)$ .

That's equivalent to

$y = 3x - 16$ .

Refer to the diagram attached. The line touches the graph of y = f(x) at x = 3 without crossing it. The line here is thus a tangent to the graph of y = f(x) at x = 3. The slope of the line represents the instantaneous rate of increase of f(x) at $x_{0} = 3$ .

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