Question

1. $y = -\frac{1}{4}x^2 - 1$$a=$______ $b=$______ $c=$______y-intercept _________vertex ( _________ _________ )equation for axis of symmetry $x=$ _______opens up or opens downmaximum or minimummax/min value _______

Explanation:

Step1: Identify a, b, c values

Compare $y = -\frac{1}{4}x^2 - 1$ to standard form $y=ax^2+bx+c$:
$a = -\frac{1}{4}$, $b = 0$, $c = -1$

Step2: Find y-intercept

Y-intercept is $c$:
$y\text{-intercept} = -1$

Step3: Calculate vertex x-coordinate

Use $x = -\frac{b}{2a}$:
$x = -\frac{0}{2\times(-\frac{1}{4})} = 0$

Step4: Calculate vertex y-coordinate

Substitute $x=0$ into the equation:
$y = -\frac{1}{4}(0)^2 - 1 = -1$
Vertex: $(0, -1)$

Step5: Find axis of symmetry

Axis of symmetry is $x =$ vertex x-coordinate:
$x = 0$

Step6: Determine opening direction

If $a<0$, parabola opens down:
$a = -\frac{1}{4} < 0$, so opens down

Step7: Identify max/min

Downward parabola has a maximum:
Maximum at vertex y-value

Step8: Find max value

Max value = vertex y-coordinate:
$\text{Max Value} = -1$

Answer:

$a = -\frac{1}{4}$, $b = 0$, $c = -1$
Y-intercept: $-1$
Vertex: $(0, -1)$
Equation for Axis of Symmetry: $x = 0$
Opens Down
Maximum
Max Value: $-1$

(To graph: Plot the vertex $(0,-1)$, y-intercept $(0,-1)$, and additional points e.g., $x=2, y=-\frac{1}{4}(4)-1=-2$; $x=-2, y=-2$, then draw a downward-opening parabola symmetric over $x=0$)