What is the sum of all numbers a for which the graph of y=x^2+a and the graph of y=ax intersect one time?

Guest Nov 12, 2019

#1**+3 **

It intersects one time if they have ONE solution.

We substitute Y.

\(x^2+a=ax\)

Then we set it equal 0 so it matches the standard quadratic form.

\(x^2-ax+a=0\)

In order to find when it has ONE solution, the discriminant of the equation ABOVE has to be equal to 0.

The discriminant is:

\(b^2-4ac\)

We plug in values.

\((-a)^2-4(a)=0\)

We have:

\(a^2-4a=0\)

We factor:

\(a(a-4)=0\)

a = 0 or a = 4

So the sum of the values is 0 + 4 = \(\boxed{4}\)

.CalculatorUser Nov 12, 2019

#1**+3 **

Best Answer

It intersects one time if they have ONE solution.

We substitute Y.

\(x^2+a=ax\)

Then we set it equal 0 so it matches the standard quadratic form.

\(x^2-ax+a=0\)

In order to find when it has ONE solution, the discriminant of the equation ABOVE has to be equal to 0.

The discriminant is:

\(b^2-4ac\)

We plug in values.

\((-a)^2-4(a)=0\)

We have:

\(a^2-4a=0\)

We factor:

\(a(a-4)=0\)

a = 0 or a = 4

So the sum of the values is 0 + 4 = \(\boxed{4}\)

CalculatorUser Nov 12, 2019