All categories

3 years ago

Which of the following are characteristics of the graph of the quadratic parent function

Mathematics

2 answers:

Stolb23 [73]3 years ago

8 0

Answer:

D. It is U-shaped

Step-by-step explanation:

The standard form of a quadratic function is $y=a*x^{2} +b*x+c$ . The graph of quadratic function opens in the direction of the positive y-axis or negative y-axis, it depends on the coefficient of $x^{2}$ in the function. The shape of the graph of quadratic function is U shaped, because the greatest power of the x in the function is 2.

Hence The shape of quadratic function is U-shaped. It is the characteristics of the quadratic parent function.

Bogdan [553]3 years ago

4 0

It would be A. because it intercepts the x-axis and y-axis at (0, 0) and D. because it looks like a u

You might be interested in

If a figure located at T(-3, -1), U(1, -2), V(2, 2), W(-1, 1) and is rotated 90∘ counterclockwise rotation about the origin, whi

damaskus [11]

Answer:

(1, - 3 )

Step-by-step explanation:

Under a counterclockwise rotation about the origin of 90°

a point (x, y ) → (- y, x ), thus

T(- 3, - 1 ) → T'(1, - 3 )

3 0

3 years ago

Fantom [35]

Answer:

the mode remains the same

Step-by-step explanation:

42 is outlier.

mod is the most repetitive number of a number string, and outlier does not change it.

the mode remains the same.

7 0

3 years ago

(x – 2) / 4 – (3x + 5) / 7 = – 3, x = ?<br><br> WILL GIVE BRAINLIEST

balu736 [363]

Answer:

under

Step-by-step explanation:

change the underminator, we have

7×(x-2)-4×(3x+5)=-3×7×4

7x-14-12x-20=-84

-5x=-50

x=10

6 0

2 years ago

-4 + 4d + 17 = 33<br> D = ?

cricket20 [7]

Answer:

Step-by-step explanation:

I think it is 3, because...

33 - 17 = 16

16 - 4 = 12

12/4 = 3

???

7 0

2 years ago

Solve 73 make sure to also define the limits in the parts a and b

Aleks04 [339]

73.

$f(x)=\frac{3x^4+3x^3-36x^2}{x^4-25x^2+144}$

$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3$ $\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\cdot\frac{1}{2}=3$

Since we can't divide by zero, we need to find when:

$x^4-2x^2+144=0$

But before, we can factor the numerator and the denominator:

$\begin{gathered} \frac{3x^2(x^2+x-12)}{x^4-25x^2+144}=\frac{3x^2((x+4)(x-3))}{(x-3)(x-3)(x+4)(x+4)} \\ so: \\ \frac{3x^2}{(x+3)(x-4)} \end{gathered}$

Now, we can conclude that the vertical asymptotes are located at:

$\begin{gathered} (x+3)(x-4)=0 \\ so: \\ x=-3 \\ x=4 \end{gathered}$

so, for x = -3:

$\lim_{x\to-3^-}f(x)=\lim_{x\to-3^-}-\frac{162}{x^4-25x^2+144}=-162(-\infty)=\infty$ $\lim_{x\to-3^+}f(x)=\lim_{x\to-3^+}-\frac{162}{x^4-25x^2+144}=-162(\infty)=-\infty$

For x = 4:

$\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty$ $\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty$

4 0

11 months ago