Question

Analyze the key features of the graphs of the functions below. Select all of the quadratic functions that open down, have a vertex that is a maximum and a positive y-intercept.
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) = 2x^2 - 4x - 3; opens up; vertex: (1, -5); y-intercept: (0, -3)

g(x) = -x^2 + x + 1; opens down; vertex: (0.5, 1.25); y-intercept: (0, 1)

h(x) = -2x^2 + 3x - 1; opens down; vertex: (0.75, 0.125); y-intercept: (0, -1)

m(x) = x^2 - 9; opens up; vertex: (0, -9); y-intercept: (0, -9)

n(x) = -3x^2 + 7; opens down; vertex: (0, 7); y-intercept: (0, 7)

I have attached an image of the functions and their graphs.

Hope this helps!

Answer 2

Answer:

$f(x) = -x^2 +x +1$

Step-by-step explanation:

Quadratic equation is in the form of

$f(x)= ax^2+bx+c$

When the value of 'a' is positive then the graph opens up

When the value of 'a' is negative  then the graph opens down

When the graph opens up then the vertex is minimum.

When the graph opens down then the vertex is maximum.

y intercept is the value of 'c'

$f(x) = 2x^2 - 4x - 3$, a=2 the graph opens up

$f(x) = -x^2 +x +1$, a=-1 the graph opens down. So the vertex is maximum. Y intercept is +1. y intercept is positive.

$f(x) = -2x^2 +3x - 1$, a=-2 the graph opens down. So the vertex is maximum. Y intercept is -1. y intercept is negative.

$f(x) = 2x^2 - 4x - 3$, a=2 the graph opens up