Determine the equation of each of the following parabolas

Mathematics

1 answer:

nata0808 [166]4 years ago

4 0

Try this solution with a short explanation:

a) y=ax²+5; ⇒ substitution (4;0): 16a+5=0; ⇒ a= -5/16.

Answer: y= -5/16 x²+5;

b) y=x²;

c) y=ax(x+7); ⇒ substitution (4;4): 4a(4+7)=4; ⇒a=1/11.

Answer: y= 1/11 x²+ 7/11 x;

d) y=a(x-1)(x-3); ⇒substitution (0;3): (-1)*(-3)*a=3; ⇒a=1.

Answer: y=x²-4x+3;

e) y=a(x-1)(x+3); ⇒ substitution (-1;5): (-2)*2*a=5; ⇒ a= -5/4.

Answer: y= -5/4 x² -5/2 x +15/4;

f) y=ax²+bx+c; ⇒ substitution (2;2): 4a+2b+6=2 or 2a+b= -2 (the I-st equation of the system);

the x-coordinate of the vertex: -b/2a=2; ⇒ b= -4a (the II-d equation of the system);

$\left \{ {{2a+b=-2} \atop {b=-4a}} \right. \ => \ \left \{ {{a=1} \atop {b=-4}} \right.$

Answer: y=x²-4x+6.

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7 0

4 years ago

(1 point) The matrix A=⎡⎣⎢−4−4−40−8−4084⎤⎦⎥A=[−400−4−88−4−44] has two real eigenvalues, one of multiplicity 11 and one of multip

serious [3.7K]

Answer:

We have the matrix $A=\left[\begin{array}{ccc}-4&-4&-4\\0&-8&-4\\0&8&4\end{array}\right]$

To find the eigenvalues of A we need find the zeros of the polynomial characteristic $p(\lambda)=det(A-\lambda I_3)$

Then

$p(\lambda)=det(\left[\begin{array}{ccc}-4-\lambda&-4&-4\\0&-8-\lambda&-4\\0&8&4-\lambda\end{array}\right] )\\=(-4-\lambda)det(\left[\begin{array}{cc}-8-\lambda&-4\\8&4-\lambda\end{array}\right] )\\=(-4-\lambda)((-8-\lambda)(4-\lambda)+32)\\=-\lambda^3-8\lambda^2-16\lambda$

Now, we fin the zeros of $p(\lambda)$ .

$p(\lambda)=-\lambda^3-8\lambda^2-16\lambda=0\\\lambda(-\lambda^2-8\lambda-16)=0\\\lambda_{1}=0\; o \; \lambda_{2,3}=\frac{8\pm\sqrt{8^2-4(-1)(-16)}}{-2}=\frac{8}{-2}=-4$

Then, the eigenvalues of A are $\lambda_{1}=0$ of multiplicity 1 and $\lambda{2}=-4$ of multiplicity 2.

Let's find the eigenspaces of A. For $\lambda_{1}=0$ : $E_0 = Null(A- 0I_3)=Null(A)$ .Then, we use row operations to find the echelon form of the matrix