The graphs of the equations y = x^2 - 10 and y = 2x - 2 intersect at the point (x, y) in t… — SAT

The graphs of the equations $y = x^2 - 10$ and $y = 2x - 2$ intersect at the point $(x, y)$ in the first quadrant. What is the value of $y$ ?

Answer: 6

The value of

y

is 6.

Equating the equations gives

x^2 - 10 = 2x - 2

. Moving all terms to one side yields

x^2 - 2x - 8 = 0

, which factors as

(x - 4)(x + 2) = 0

. Since the point is in the first quadrant, both coordinates must be positive, so we use

x = 4

. Substituting

x = 4

into the linear equation gives

y = 2(4) - 2 = 6

Equate the two expressions for

y

x^2 - 10 = 2x - 2

Since both equations define

y

, their right-hand sides must be equal at the points of intersection.

Rewrite the equation in standard quadratic form

ax^2 + bx + c = 0

x^2 - 2x - 8 = 0

Subtract

2x

and add

2

to both sides of the equation to set it equal to zero.

Factor the quadratic equation

(x - 4)(x + 2) = 0

Find two integers that multiply to

-8

and add to

-2

, which are

-4

and

2

Solve for the possible values of

x

x = 4

x = -2

Set each factor equal to zero and solve.

Determine the positive

x

-coordinate and find

y

y = 6

For the point to be in the first quadrant, both coordinates must be positive. Thus, we select

x = 4

and substitute it into the linear equation:

y = 2(4) - 2 = 6

Solving a system of linear and quadratic equations via substitution.

Instead of factoring, the quadratic formula can be used to solve

x^2 - 2x - 8 = 0

x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-8)}}{2(1)} = \frac{2 \pm \sqrt{36}}{2} = \frac{2 \pm 6}{2}

. This yields

x = 4

and

x = -2

. Then substitute the positive root to find

y

Estimated Time:1m 30s

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