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quadratic function

Authored by Agnieszka Jankowska

Mathematics

10th Grade

Used 5+ times

quadratic function
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10 questions

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1.

MULTIPLE SELECT QUESTION

3 mins • 1 pt

The set:

 <2,+ <-2,+\infty\   )  is the range of:

 f(x)=2(x3)22f\left(x\right)=-2\left(x-3\right)^2-2  

 f(x)=2(x3)22f\left(x\right)=2\left(x-3\right)^2-2  

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

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

2.

MULTIPLE SELECT QUESTION

3 mins • 1 pt

 The quadratic equation

 x2cx+4=0,  cRx^2-cx+4=0,\ \ c\in R  has exactly one solution if and only if:

 c4c\ne4  

 cR{4,4}c\in R-\left\{-4,4\right\}  

 c=4c=4  

 c{4,4}c\in\left\{-4,4\right\}  

3.

FILL IN THE BLANK QUESTION

3 mins • 1 pt

The smallest value of the function    f(x)=2(x3)(x5) f\left(x\right)=2\left(x-3\right)\left(x-5\right)\    in the interval  <6,4><-6,4>  is:

4.

MULTIPLE SELECT QUESTION

3 mins • 1 pt

Zeros of a quadratic function   y=f(x)y=f\left(x\right)  are numbers:  2,42,-4  and its graph goes through the point  P=(2,8)P=\left(-2,-8\right) . Its formula is:

 f(x)=2(x+1)29f\left(x\right)=2\left(x+1\right)^2-9  

 f(x)=2(x2)2+8f\left(x\right)=2\left(x-2\right)^2+8  

 f(x)=(x2)28f\left(x\right)=-\left(x-2\right)^2-8  

 f(x)=(x+1)29f\left(x\right)=\left(x+1\right)^2-9  

5.

MULTIPLE SELECT QUESTION

3 mins • 1 pt

Media Image

The graph of the function   f(x)=ax2+bx+c,   a0f\left(x\right)=ax^2+bx+c,\ \ \ a\ne0  is thown on the picture.  Then:

 a<0, b<0, c>0a<0,\ b<0,\ c>0  

 a>0, b<0, c>0a>0,\ b<0,\ c>0  

 a<0, b>0, c>0a<0,\ b>0,\ c>0  

 a<0, b>0, c<0a<0,\ b>0,\ c<0  

6.

MULTIPLE SELECT QUESTION

3 mins • 1 pt

The quadratic function:   f(x)=2x2+x5 f\left(x\right)=2x^2+x-5\   has two zeros:  x1,  x2x_1,\ \ x_2  . Then:

 x12+x22=5,25x_1^2+x_2^2=5,25  

 x12+x22=2413x_1^2+x_2^2=\frac{2\sqrt{41}}{3}  

 x12+x22=1241x_1^2+x_2^2=12-\sqrt{41}  

 x12+x22=4,5x_1^2+x_2^2=4,5  

7.

MULTIPLE SELECT QUESTION

3 mins • 1 pt

The equation   x2cx+9=0x^2-cx+9=0  has no solutions if and only if:

 c(,6)(6,+)c\in\left(-\infty,-6\right)\cup\left(6,+\infty\right)  

 c(6,6)c\in\left(-6,6\right)  

 c(,6><6,+)c\in\left(-\infty,-6>\cup<6,+\infty\right)  

 c<6,6>c\in<-6,6>  

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