QuestionFind where the lines $y = x - 3$ and $y = 3x^2$ intersect.

Studdy Solution

STEP 1

Assumptions1. The first equation is $y = x -3$ . The second equation is y =3x^
3. We are looking for the points of intersection, which are the points $(x, y)$ where both equations are true at the same time.

STEP 2

To find the points of intersection, we can set the two equations equal to each other and solve for $x$ .
$x - =x^2$

STEP 3

Rearrange the equation to form a quadratic equation.
$3x^2 - x +3 =0$

STEP 4

This is a quadratic equation in the form of $ax^2 + bx + c =0$ . We can solve it using the quadratic formula, which is $x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$ .

STEP 5

Plug in the values for $a$ , $b$ , and $c$ into the quadratic formula.
$x = \frac{-(-1) \pm \sqrt{(-1)^2 -433}}{2*3}$

STEP 6

implify the equation.
$x = \frac{1 \pm \sqrt{1 -36}}{6}$

STEP 7

Calculate the value under the square root.
$x = \frac{1 \pm \sqrt{-35}}{6}$

STEP 8

The value under the square root is negative, which means there are no real solutions for $x$ . Therefore, the two graphs do not intersect at any real points.

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