Question

F(x) = -4x(2)+ 12x – 9

Answer 1

For this case we have the following function:

$f (x) = - 4x ^ 2 + 12x-9$

$y = 0$ we have:

$-4x ^ 2 + 12x-9 = 0$

Where:

$a = -4\\b = 12\\c = -9$

By definition, the discriminant of a quadratic equation is given by:

$d = b ^ 2-4 (a) (c)$

$d> 0:$ Two different real roots

$d = 0:$ Two equal real roots

$d$: Two different complex roots

Substituting the values we have:

$d = (12) ^ 2-4 (-4) (- 9)\\d = 144-144$

$d = 0$

We have two equal real roots.

To find the intersections with the x axis, we do $y = 0:$

$-4x ^ 2 + 12x-9 = 0$

We apply the quadratic formula:

$x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}$

Substituting the values we have:

$x = \frac {-12 \pm \sqrt {12 ^ 2-4 (-4) (- 9)}} {2 (-4)}\\x = \frac {-12 \pm \sqrt {144-144}} {- 8}\\x = \frac {-12 \pm0} {- 8}\\x = \frac {-12} {- 8}\\x = \frac {3} {2}$

The intersection with the x axis is$(\frac {3} {2}, 0)$

Answer:

$d = 0$

The intersection with the x axis is $(\frac {3} {2}, 0)$

Answer 2

Answer:

Part A) The value of the discriminant of f is zero

Part B) The quadratic equation has only one x-intercept

Step-by-step explanation:

Part A) What is the value of the discriminant of f?

we know that

In a quadratic equation of the form

$ax^{2} +bx+c=0$

the discriminant is equal to

$D=b^2-4ac$

in this problem we have

$f(x)=-4x^2+12x-9$

so

$a=-4\\b=12\\c=-9$

substitute

$D=12^2-4(-4)(-9)$

$D=0$

Part B) How many x-intercepts does the graph of f have?

we know that

In a quadratic equation

If the discriminant D is equal to zero, then the equation has only one real solution

That means

The quadratic equation has only one x-intercept