Given is the graph of f(x)= ax^2+bx+c (a?0). Then match the items in column 1 to coulmn 2.
Please solve this and explain each step
Dear Student,
In graph P, a>0 (since the parabola opens upwards), c>0 (as the y intercept is positive), b>0 (since x intercept are positives). As, we have positive unequal roots, so D > 0. So, for P option (A) is correct.
In graph Q, a>0 (since the parabola opens downwards), c>0 (as the y intercept is positive), b>0 (since x intercept are positives). As, we have positive unequal roots, so D > 0. So, for P option (B) is correct.
In graph R, a>0 (since the parabola opens upwards), c>0 (as the y intercept is positive), b>0 (since vertex lies in the positive side of x axis). So, for P option (C) is correct.
In graph S, a<0 (since the parabola opens downwards), c<0 (as the y intercept is negative), b>0 (since vertex lies in the positive side of x axis ). So, for P option (C) is correct.
Option (d) is not true for any.
Regards,
In graph P, a>0 (since the parabola opens upwards), c>0 (as the y intercept is positive), b>0 (since x intercept are positives). As, we have positive unequal roots, so D > 0. So, for P option (A) is correct.
In graph Q, a>0 (since the parabola opens downwards), c>0 (as the y intercept is positive), b>0 (since x intercept are positives). As, we have positive unequal roots, so D > 0. So, for P option (B) is correct.
In graph R, a>0 (since the parabola opens upwards), c>0 (as the y intercept is positive), b>0 (since vertex lies in the positive side of x axis). So, for P option (C) is correct.
In graph S, a<0 (since the parabola opens downwards), c<0 (as the y intercept is negative), b>0 (since vertex lies in the positive side of x axis ). So, for P option (C) is correct.
Option (d) is not true for any.
Regards,