The line y=9x-4 intersects the quadratic function y=x^2+7x at one point. What are the coordinates of the point of intersection?
that would be where
9x-4 = x^2+7x
x^2-2x+4 = 0
That has no real solutions, so I suspect a typo. See that the graphs do not intersect:
http://www.wolframalpha.com/input/?i=plot+y%3D9x-4+%2C+y+%3D+x^2%2B7x
Now, the line y=9x+4 does intersect the parabola.
To find the point of intersection between the line y=9x-4 and the quadratic function y=x^2+7x, we need to set the two equations equal to each other and solve for x.
Setting the equations equal, we get:
9x-4 = x^2+7x
Rearranging the equation, we have:
x^2+7x - 9x + 4 = 0
Combining like terms, we get:
x^2 - 2x + 4 = 0
Now, we need to solve this quadratic equation to find the values of x. To do this, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 1, b = -2, and c = 4.
Plugging these values into the quadratic formula, we have:
x = (-(-2) ± √((-2)^2 - 4(1)(4))) / (2 * 1)
x = (2 ± √(4 - 16)) / 2
x = (2 ± √(-12)) / 2
Since the square root of a negative number is not a real number, it means there are no real solutions for x. Therefore, the line y=9x-4 does not intersect the quadratic function y=x^2+7x at any point.
To find the point of intersection between the line y=9x-4 and the quadratic function y=x^2+7x, we need to set the two equations equal to each other and solve for x.
The equation for the line is: y = 9x - 4
The quadratic function is: y = x^2 + 7x
Setting the two equations equal to each other:
9x - 4 = x^2 + 7x
Rearranging the equation to form a quadratic equation equal to zero:
x^2 - 2x - 4 = 0
To solve this quadratic equation, we can either factor or use the quadratic formula. In this case, factoring may not be possible, so we'll use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
For our equation, a = 1, b = -2, and c = -4. Substituting these values into the formula:
x = (-(-2) ± √((-2)^2 - 4(1)(-4))) / (2(1))
Simplifying the formula:
x = (2 ± √(4 + 16)) / 2
x = (2 ± √20) / 2
x = (2 ± 2√5) / 2
Simplifying further:
x = 1 ± √5
So we have two possible x-values for the point of intersection: x = 1 + √5 and x = 1 - √5.
To find the corresponding y-values, we can substitute these x-values into either of the original equations. Let's use the quadratic function:
For x = 1 + √5:
y = (1 + √5)^2 + 7(1 + √5)
= 1 + 2√5 + 5 + 7 + 7√5
= 13 + 9√5
For x = 1 - √5:
y = (1 - √5)^2 + 7(1 - √5)
= 1 - 2√5 + 5 - 7 - 7√5
= -1 - 9√5
Therefore, the coordinates of the point of intersection are (1 + √5, 13 + 9√5) and (1 - √5, -1 - 9√5).