 William Curry

2021-12-11

Determine the vertex and axis of symmetry of the following quadratics.
a) $f\left(x\right)=2\left(x-3\right)2-1$
b) $f\left(x\right)={x}^{2}+4x+3$ Step 1
For a quadratic equation $y=a{x}^{2}+bx+c$
Axis of Symmetry is $x=\frac{-b}{2a}$
x-coordinate of vertex is $\frac{-b}{2a}$
a) $f\left(x\right)=2{\left(x-3\right)}^{2}-1$
$=2\left({x}^{2}-6x+9\right)-1$
$f\left(x\right)=2{x}^{2}-12x+17$
Axis of symmetry $⇒x=\frac{-b}{2a}$
$⇒x=-\frac{\left(-12\right)}{2\left(2\right)}=3$
$⇒x=3$
Step 2
x-coordinate of vertex $=\frac{-b}{2a}=3$
y-coordinate of vertex $=f\left(\frac{-b}{2a}\right)=f\left(3\right)$
$=2{\left(3-3\right)}^{2}-1=-1$
$\therefore$ Vertex
b) $f\left(x\right)={x}^{2}+4x+3$
Axis of symmetry
$⇒x=\frac{-b}{2a}=\frac{-4}{2\left(1\right)}=-2$
$⇒x=-2$
x-coordinate of vertex $=\frac{-b}{2a}=-2$
Step 3
y-coordinate of vertex
$=f\left(\frac{-b}{2a}\right)$
$=f\left(-2\right)$
$={\left(-2\right)}^{2}+4\left(-2\right)+3$
$=-1$
$\therefore$ Vertex Donald Cheek

Step 1
$f\left(x\right)=2\left(x-3\right)2-1$
Multiply 2 and 2 to get 4.
$4\left(x-3\right)-1$
Use the distributive property to multiply 4 by $x-3$.
$4x-12-1$
Subtract 1 from -12 to get -13
$4x-13$
Step 2
$f\left(x\right)={x}^{2}+4x+3$
Factor the expression by grouping. First, the expression needs to be rewritten as ${x}^{2}+ax+bx+3$. To find a and b, set up a system to be solved
$a+b=4$
$ab=1×3=3$
Since ab is positive, a and b have the same sign. Since $a+b$ is positive, a and b are both positive. The only such pair is the system solution.
$a=1$
$b=3$
Rewrite ${x}^{2}+4x+3$ as $\left({x}^{2}+x\right)+\left(3x+3\right)$
$\left({x}^{2}+x\right)+\left(3x+3\right)$
Factor out x in the first and 3 in the second group.
$x\left(x+1\right)+3\left(x+1\right)$
Factor out common term $x+1$ by using distributive property.
$\left(x+1\right)\left(x+3\right)$

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