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Analyze the following quadratic functions. Choose the option(s) that open up, have a vertex that is a minimum and a negative y-intercept. Select all that apply.
F(x) = 2x2 - 4x - 3
g(x) = x2 + x + 1 h(x) = - 2x2 + 3x - 1
m(x) = x2 - 9
n(x) = - 3x2 + 7

Respuesta :

f(x) and m(x) is a vertex with a minimum because both have the negative intercepts of -3 and -9 and there is no reflection in the x-axis like the others. If there was a reflection in the x-axis there would be maximums.

Answer:

[tex]F(x) = 2x^2 - 4x - 3[/tex]

[tex]m(x) = x^2 - 9[/tex]

Step-by-step explanation:

Since, the general form of a quadratic function is,

[tex]y=ax^2+bx+c[/tex]

When, a > 0 then the graph opens up or have a vertex that is a minimum,

And, when a < 0 then the graph is opens down or have a vertex that is maximum,

Now, If the c > 0 then the y-intercept of the function is positive,

While, if c < 0 then the y-intercept of the function is negative.

[tex]F(x) = 2x^2 - 4x - 3[/tex]

2 > 0 ⇒ f(x) opens up,

Also, -3 < 0, ⇒ f(x) has a negative y-intercept.

[tex]g(x) = x^2 + x + 1[/tex]

1 > 0 ⇒ g(x) opens up,

Also, 1 >0, ⇒ g(x) has a positive y-intercept.

[tex]h(x) = - 2x^2 + 3x - 1 [/tex]

- 2 < 0 ⇒ h(x) opens down,

Also, -1 < 0, ⇒ h(x) has a negative y-intercept.

[tex]m(x) = x^2 - 9 [/tex]

1 > 0 ⇒ m(x) opens up,

Also, -9 < 0, ⇒ m(x) has a negative y-intercept.

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