Determine $p$ such that $f(x)=-3x^2+7$ and $g(x)=4x-p$ have two points of intersection.
Antwoord 1 correct
Correct
Antwoord 2 optie
$p>-5\frac{2}{3}$
Antwoord 2 correct
Fout
Antwoord 3 optie
$p<-8\frac{1}{3}$
Antwoord 3 correct
Fout
Antwoord 4 optie
$p<-5\frac{2}{3}$
Antwoord 4 correct
Fout
Antwoord 1 optie
$p>-8\frac{1}{3}$
Antwoord 1 feedback
Correct: $-3x^2+7=4x-p \Leftrightarrow -3x^2-4x+7+p=0$.
Two points of intersection implies that $D>0$. $D=(-4)^2-4\cdot-3\cdot(7+p)=100+12p$. Hence, $D>0$ if $p>-8\frac{1}{3}$.
Go on.
Two points of intersection implies that $D>0$. $D=(-4)^2-4\cdot-3\cdot(7+p)=100+12p$. Hence, $D>0$ if $p>-8\frac{1}{3}$.
Go on.
Antwoord 2 feedback
Wrong: Pay attention to the signs of the coefficients $a$, $b$ and $c$ of the quadratic function of which you determine the zeros.
Try again.
Try again.
Antwoord 3 feedback
Wrong: Note that for two points of intersection $D>0$ is required of the equation that is put to zero.
See Extra explanation: zeros.
See Extra explanation: zeros.
Antwoord 4 feedback
Wrong: Note that for two points of intersection $D>0$ is required of the equation that is put to zero.
See Extra explanation: zeros.
See Extra explanation: zeros.