Question

f(x)=x^2+6x-27

a. vertex.
b. axis of symmetry.
c. intercepts.
d. graph.
e. interval f is increasing.
f. interval f is decreasing.

Answer 1

\(\bf \textit{vertex of a parabola}\\ \quad \\
f(x) = {\color{red}{ 1}}x^2{\color{blue}{ +6}}x{\color{green}{ -27}} \quad
\left(-\cfrac{{\color{blue}{ b}}}{2{\color{red}{ a}}}\quad ,\quad {\color{green}{ c}}-\cfrac{{\color{blue}{ b}}^2}{4{\color{red}{ a}}}\right)\)

axis of symmetry, well, use the coordinates from the vertex

intercepts, set f(x) = 0, that is \(\bf 0=x^2+6x-27\) solve for "x" like you'd any quadratic, by factoring

graph, well, use the vertex, pick a point to the left of it, and one to the right of it and graph away

where is it increasing or decreasing? well, look at the graph :)

Answer 2

vertex: (-3,25.5) ? @jdoe0001

Answer 3

well.... not quite

Answer 4

oh...

Answer 5

please explain?

Answer 6

\(\bf \textit{vertex of a parabola}\\ \quad \\
f(x) = {\color{red}{ 1}}x^2{\color{blue}{ +6}}x{\color{green}{ -27}} \quad
\left(-\cfrac{{\color{blue}{ 6}}}{2{\color{red}{ (1)}}}\quad ,\quad {\color{green}{ -27}}-\cfrac{{\color{blue}{ 6}}^2}{4{\color{red}{ (1)}}}\right)\)

Answer 7

ooooh i didnt see the 2 by the six

Answer 8

(-3,-36) ?

Answer 9

well... ahemm \(\bf \textit{vertex of a parabola}\\ \quad \\
f(x) = {\color{red}{ 1}}x^2{\color{blue}{ +6}}x{\color{green}{ -27}} \quad
\left(-\cfrac{{\color{blue}{ 6}}}{2{\color{red}{ (1)}}}\quad ,\quad {\color{green}{ -27}}-\cfrac{{\color{blue}{ 6}}^2}{4{\color{red}{ (1)}}}\right)\\ \quad \\\implies \left(-3\quad ,\quad -27-\cfrac{36}{4}\right)
\implies (-3\quad ,\quad -27-9)\)

Answer 10

ya and then you subtract 9 from -27 dont you?

Answer 11

ohh yea... yes you do ... ahemm right.. so -36 correct

Answer 12

the parabola is "x" based, meaning is vertically opening
the leading term coefficient is positive, so it's going upwards