Question

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

Answer 1

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 2

Answer:

$F(x) = 2x^2 - 4x - 3$

$m(x) = x^2 - 9$

Step-by-step explanation:

Since, the general form of a quadratic function is,

$y=ax^2+bx+c$

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.

$F(x) = 2x^2 - 4x - 3$

2 > 0 ⇒ f(x) opens up,

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

$g(x) = x^2 + x + 1$

1 > 0 ⇒ g(x) opens up,

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

$h(x) = - 2x^2 + 3x - 1$

- 2 < 0 ⇒ h(x) opens down,

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

$m(x) = x^2 - 9$

1 > 0 ⇒ m(x) opens up,

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