Quadratics 1

If x = 7 and x = -8 are the roots of a quadratic function, then the equation of its axis of symmetry is

• $x=1$

• $x=-0.5$
  

• $x=0.5$
  

• $x=7.5$
  

 $f\left(x\right)=2x^2-16x+31$ 

 The vertex form is

•  $f\left(x\right)=2\left(x-8\right)^{2\ }+63$  

•  $f\left(x\right)=\left(x+4\right)^{2\ }-50$  

•  $f\left(x\right)=2\left(x+4\right)^2-1$  

•  $f\left(x\right)=2\left(x-4\right)^2-1$  

 $f\left(x\right)=2x\left(x-6\right)$  

The vertex is

•  $\left(0.6\right)$  

•  $\left(3,-18\right)$  

•  $\left(6,0\right)$  

•  $\left(-3,-18\right)$  

Which quadratic function opens upwards and has a vertex at ( 0 , 3 )?

• $f\left(x\right)=-\left(x-3\right)^2$

• $f\left(x\right)=-2x^{2\ }+3$

• $f\left(x\right)=2\left(x-3\right)^2$

• $f\left(x\right)=2x^{2\ }+3$

 $f\left(x\right)=x^2+2x-8$ 

The factored form is

•  $f\left(x\right)=\left(x-4\right)\left(x+2\right)$  

•  $f\left(x\right)=2x\left(x-8\right)$  

•  $f\left(x\right)=\left(x-4\right)\left(x-2\right)$  

•  $f\left(x\right)=\left(x+4\right)\left(x-2\right)$  

 $f\left(x\right)=-3\left(x-4\right)\left(x+5\right)$ 

The y-intercept is

•  $y=-60$  

•  $y=60$  

•  $y=-20$  

•  $y=12$  

 $f\left(x\right)=6x^{2\ }-4x+8$ 

The factored form is

•  $f\left(x\right)=6x\left(3x-2\right)$  

•  $f\left(x\right)=\left(6x-1\right)\left(3x+2\right)$  

•  $f\left(x\right)=2\left(3x^2-2x+4\right)$  

•  $f\left(x\right)=2\left(3x^2-4x-8\right)$  

 $f\left(x\right)=\frac{1}{2}\left(x-5\right)^2+1$ 

The standard form is

•  $f\left(x\right)=2x^2-10x\ +2$  

•  $f\left(x\right)=\frac{1}{2}x^{2\ }+25x+24$  

•  $f\left(x\right)=\frac{1}{2}x^{2\ }-5x+\frac{27}{2}$  

•  $f\left(x\right)=2x^{2\ }-10x+27$  

 $f\left(x\right)=3x^2-4x+11$ 

The parabola

• opens up and has the y-intercept at y = 11
  

• opens down and has the y-intercept at y = 11
  

• has the y-intercept at y = 11/3

• opens up and has no y-intercept
  

The vertex form is

• $f\left(x\right)=-\left(x-3\right)^2+2$

• $f\left(x\right)=-3\left(x-2\right)^{2\ }+3$

• $f\left(x\right)=3\left(x-2\right)^2+3$

• $f\left(x\right)=3\left(x-2\right)^{2\ }-3$