How to graph a quadratic function - Geometry

is a polynomial function, and as such has the set of all real numbers as its domain.

is a polynomial function, and as such has the set of all real numbers as its domain.

is a polynomial function, and as such has the set of all real numbers as its domain.

The -intercepts, if any exist, can be found by setting :

The only -intercept is .

The -intercept can be found by substituting 0 for :

The -intercept is .

The correct set of intercepts is .

is a quadratic function.

One way to find the maximum or minimum value of a quadratic function such as is to find the vertex of its parabola; its -coordinate will be the one extremum of the range. Whether it is the minimum or maximum value depends on the quadratic coefficient; since it is a positive value here (1), it will be a minimum value.

Set and equal to quadratic and linear coefficients 1 and 4, respectively, and evaluating the expression :

This gives the -coordinate of the vertex; the -coordinate, and the minimum value of , can be found by evaluating , which is done by substitution:

The range is therefore .

The -intercepts, if any exist, can be found by setting :

The only -intercept is .

The -intercept can be found by substituting 0 for :

The -intercept is .

The correct set of intercepts is .

is a quadratic function.

One way to find the maximum or minimum value of a quadratic function such as is to find the vertex of its parabola; its -coordinate will be the one extremum of the range. Whether it is the minimum or maximum value depends on the quadratic coefficient; since it is a positive value here (1), it will be a minimum value.

Set and equal to quadratic and linear coefficients 1 and 4, respectively, and evaluating the expression :

This gives the -coordinate of the vertex; the -coordinate, and the minimum value of , can be found by evaluating , which is done by substitution:

The range is therefore .

The -intercepts, if any exist, can be found by setting :

The only -intercept is .

The -intercept can be found by substituting 0 for :

The -intercept is .

The correct set of intercepts is .

is a quadratic function.

One way to find the maximum or minimum value of a quadratic function such as is to find the vertex of its parabola; its -coordinate will be the one extremum of the range. Whether it is the minimum or maximum value depends on the quadratic coefficient; since it is a positive value here (1), it will be a minimum value.

Set and equal to quadratic and linear coefficients 1 and 4, respectively, and evaluating the expression :

This gives the -coordinate of the vertex; the -coordinate, and the minimum value of , can be found by evaluating , which is done by substitution:

The range is therefore .

The -intercepts, if any exist, can be found by setting :

The only -intercept is .

The -intercept can be found by substituting 0 for :

The -intercept is .

The correct set of intercepts is .

is a quadratic function.

One way to find the maximum or minimum value of a quadratic function such as is to find the vertex of its parabola; its -coordinate will be the one extremum of the range. Whether it is the minimum or maximum value depends on the quadratic coefficient; since it is a positive value here (1), it will be a minimum value.

Set and equal to quadratic and linear coefficients 1 and 4, respectively, and evaluating the expression :

This gives the -coordinate of the vertex; the -coordinate, and the minimum value of , can be found by evaluating , which is done by substitution:

The range is therefore .

The line of symmetry of a vertical parabola is a vertical line, the equation of which takes the form for some . The line of symmetry passes through the vertex, which here is , so the equation must be .

If the graph of is concave upward, then .

If the graph does not intersect the -axis, then has no real solution, and the discriminant is negative:

For the parabola to have both characteristics, it must be true that and , but these two events are mutually exclusive. Therefore, the parabola cannot exist.

The line of symmetry of a horizontal parabola is a horizontal line, the equation of which takes the form for some . The line of symmetry passes through the vertex, which here is , so the equation must be .

A parabola with its -intercepts at and at has as its equation

for some nonzero . If this is multiplied out, the equation can be rewritten as

or, simplified,

The sign of quadratic coefficient determines whether it is concave upward or concave downward. We do not have the sign or any way of determining it.

The -coordinate of the -intercept is the contant, , but without knowing , we have no way of knowing .

The -coordinate of the vertex of is the value . since , this expression becomes

The -coordinate is

,

but without knowing , this coordinate, and the vertex itself, cannot be determined.

The line of symmetry is the line ; this value was computed to be equal to 6, so the line can be determined to be .

This can be answered rewriting this expression in the form

.

Once this is done, we can identify the vertex as the point .

The vertex is

I) The orientation of the parabola is determined solely by the value of . Since , the parabola can be determined to open upward.

II and V) The -coordinate of the vertex is ; since you are not given , you cannot find this. Also, since the line of symmetry has equation , for the same reason, you cannot find this either.

III) The -intercept is the point at which ; by substitution, it can be found to be at . is unknown, so the -intercept cannot be found.

IV) The -intercept(s), if any, are the point(s) at which . This is solvable using the quadratic formula

Since all three of and must be known for this to be evaluated, and only is known, the -intercept(s) cannot be identified.

The correct response is I only.

I) The orientation of the parabola is determined solely by the sign of . Since , the parabola can be determined to be concave downward.

II and V) The -coordinate of the vertex is ; since you are not given , you cannot find this. Also, since the line of symmetry has equation , for the same reason, you cannot find this either.

III) The -intercept is the point at which ; by substitution, it can be found to be at . known to be equal to 9, so the -intercept can be determined to be .

IV) The -intercept(s), if any, are the point(s) at which . This is solvable using the quadratic formula

Since all three of and must be known for this to be evaluated, and only is known, the -intercept(s) cannot be identified.

The correct response is I and III only.