Answer
a.
$-4(x+\frac{3}{2})^{2} + 10$
b.
$Vertex: (\frac{-3}{2},10)$
$x-intercepts: x= \frac{-3-\sqrt 10}{2}$ and $x= \frac{-3+\sqrt 10}{2}$
$y-intercept: f(0) = -4\times0^{2} - 12\times0 + 1 = 1$
c. See the graph
d. Domain is all real numbers and Range is from (-infinity, 10]
Work Step by Step
a.
$-4x^{2} - 12x + 1 = -4(x^{2} + 3x + \frac{9}{4}) – \frac{9}{4}\times(-4) + 1 = -4(x+\frac{3}{2})^{2} + 10$
b.
$Vertex: (\frac{-3}{2},10)$
$x-intercepts: f(x)=0$
$-4x^{2} - 12x + 1 = 0$
=> $x= \frac{-(-12) +\sqrt ((-12)^{2}-4\times(-4)\times1)}{2\times(-4)}$ and $x= \frac{-(-12) -\sqrt ((-12)^{2}-4\times(-4)\times1)}{2\times(-4)}$
=> $x= \frac{12 +\sqrt (144+16)}{-8}$ and $x= \frac{12 -\sqrt (144+16)}{-8}$
=> $x= \frac{-3-\sqrt 10}{2}$ and $x= \frac{-3+\sqrt 10}{2}$
$y-intercept: f(0) = -4\times0^{2} - 12\times0 + 1 = 1$
c. See the graph
d. From the graph
Domain is all real numbers
Range is from (-infinity, 10]