# SAT Math Multiple Choice Question 611: Answer and Explanation

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**Question: 611**

**11.** The graph of the equation *y* = 2*x*^{2} - 16*x* + 14 intersects the *y*-axis at point *A* and the *x*-axis at points *B* and *C*. What is the area of triangle *ABC*?

- A. 42
- B. 48
- C. 54
- D. 56

**Correct Answer:** A

**Explanation:**

**A Advanced Mathematics (triangles/quadratics) MEDIUM-HARD**

Any point that intersects the *y*-axis has an *x*-value of 0. So, to find point *A*, plug in 0 for *x* and solve for *y*:

*y* = 2*x*^{2} - 16*x* + 14

Plug in 0 for *x*:

*y* = 2(0)^{2} - 16(0) + 14 = 14

Any point that intersects the *x*-axis has a *y*-value of 0. So, to find points *B* and *C*, plug in 0 for *y* and solve for *x*:

*y* = 2*x*^{2} - 16*x* + 14

Substitute 0 for *y*:

0 = 2*x*^{2} - 16*x* + 14

Divide by 2:

0 = *x*^{2} - 8*x* + 7

Factor:

0 = (*x* - 7)(*x* - 1)

Use the Zero Product Property:

*x* = 7 and *x* = 1

If we connect these three points, we get a triangle with a height of 14 (from *y* = 0 to *y* = 14) and a base of 6 (from *x* = 1 to *x* = 7).

Use the triangle area formula *bh*: