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I need the explanation to understand pleasey= 5 (x + 3)^2 - 4A.) Identify the coefficients (a, h, and k) B.) Tell whether the graph opens up or opens down. C.) Find the vertex. Write as a coordinate. D.) Find the axis of symmetry. Write as an equation. E.) Find two more points on the graph. You can choose what x - values to use. Write your points as coordinates (x, y).

Sagot :

[tex]y=5(x+3)^2-4[/tex]

A)

the general form of the parabola is

[tex]y=a(x-h)^2-k[/tex]

so a=5, h=-3, k=4

B)

the graph opens depending on the sign of a, if is positive the graph opens up, and this is the case

C)

the vertex is (h,k) ever, so on this case is

[tex](-3,4)[/tex]

D)

the axis of symmetry is h ever, so on this case is -3

E)

we can give 2 values for x, for example

x=0

[tex]\begin{gathered} y=5((0)+3)^2-4 \\ y=5(3)^2-4 \\ y=41 \end{gathered}[/tex]

the first point is (0,41)

x=1

[tex]\begin{gathered} y=5((1)+3)^2-4 \\ y=5(4)^2-4 \\ y=76 \end{gathered}[/tex]

and the second point is (1,76)