Question

Please help Algebra
If the value of a in the quadratic function f(x)=ax^2+bx+c is 1/2 the function will:
open down and have a minimum
open up and have a maximum
open down and have a maximum
open up and have a minimum

Answer 1

Question

Which is correct for $f(x)=\dfrac{1}{2} x^2+bx+c$?

1. Opens down and have a minimum
2. Opens up and have a maximum
3. Opens down and have a maximum
4. Opens up and have a minimum

Solution

The quadratic functions are parabola, which is open.

If the highest degree is positive, the graph opens up and has a minimum.

Conclusion

The correct option is the 4th choice.

More Information

If the signs of coefficients of $x^2$ and $x$ is equal, the symmetry axis of the graph is on the left of the y-axis. The reason for this is because the axis of symmetry is $x=-\dfrac{b}{2a}$.

If the sign of constant term is positive, the y-intercept is positive. The reason for this is because the y-intercept is the value of $f(0)$.

If the discriminant is negative, the quadratic graph will not touch the x-axis. The reason for this is because the graph touches x-axis when $f(x)=0$, but the solution is imaginary hence it doesn't touch on the graph.

Answer 2

The quadratic functions are parabola, which is open.

If the highest degree is positive, the graph opens up and has a minimum.

The correct option is the 4th choice.

In algebra, a quadratic function, a quadratic polynomial, a polynomial of degree 2, or simply a quadratic, is a polynomial function with one or more variables in which the highest-degree term is of the second degree.

Answer-

If the signs of coefficients of $x^2$ and $x$ is equal, the symmetry axis of the graph is on the left of the y-axis. The reason for this is because the axis of symmetry is $x=-\frac{b}{2a}$.

If the sign of constant term is positive, the y-intercept is positive. The reason for this is because the y-intercept is the value of $f(0)$

If the discriminant is negative, the quadratic graph will not touch the x-axis. The reason for this is because the graph touches x-axis when $f(x)=0$ , but the solution is imaginary hence it doesn't touch on the graph.

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