To determine the intercepts of the quadratic equation [tex]\( y = -4x^2 + 2x - 1 \)[/tex], we follow these steps:
### Finding the [tex]\( y \)[/tex]-Intercept:
The [tex]\( y \)[/tex]-intercept occurs where [tex]\( x = 0 \)[/tex]. Substituting [tex]\( x = 0 \)[/tex] into the equation gives:
[tex]\[ y = -4(0)^2 + 2(0) - 1 = -1 \][/tex]
Therefore, the [tex]\( y \)[/tex]-intercept is:
[tex]\[ y = -1 \][/tex]
Choice:
A. [tex]\( y = -1 \)[/tex]
### Finding the [tex]\( x \)[/tex]-Intercept(s):
The [tex]\( x \)[/tex]-intercepts occur where [tex]\( y = 0 \)[/tex]. We set the equation to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ 0 = -4x^2 + 2x - 1 \][/tex]
This is a quadratic equation in the standard form [tex]\( ax^2 + bx + c = 0 \)[/tex] with [tex]\( a = -4 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -1 \)[/tex].
To find the [tex]\( x \)[/tex]-intercepts, we calculate the discriminant [tex]\( \Delta \)[/tex] of the quadratic formula:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
[tex]\[ \Delta = (2)^2 - 4(-4)(-1) \][/tex]
[tex]\[ \Delta = 4 - 16 \][/tex]
[tex]\[ \Delta = -12 \][/tex]
Since the discriminant is negative ([tex]\( \Delta < 0 \)[/tex]), there are no real solutions for [tex]\( x \)[/tex]. Therefore, there are no [tex]\( x \)[/tex]-intercepts.
Choice:
B. There is no [tex]\( x \)[/tex]-intercept.
In summary:
- The [tex]\( y \)[/tex]-intercept is [tex]\( y = -1 \)[/tex].
- There are no [tex]\( x \)[/tex]-intercepts.
