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**Finding x- and y-intercepts algebraically**

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y-intercepts For every point along the y-axis, the value of x will be zero (x=0) We can use this fact to find y-intercepts if we have an equation for a function. EXAMPLE: If y = 2x – 3 and we know that x = 0… we can plug in zero for x and simplify to find y: y = 2(0) – 3 y = 0 – 3 y = -3 This means that if we were to graph f(x) = 2x – 3, as you can see on the next slide, the graph would cross the y-axis at the point (0, -13) where x = 0 and y = -13

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Graph of f(x) = y = 2x – 3

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x-intercepts For every point along the x-axis, the value of y will be zero (x=0) We can use this fact to find x-intercepts if we have an equation for a function. EXAMPLE: If y = -½x2 + 8 and we know that y = 0… we can plug in zero for y and solve to find x: 0 = -½x x2 = 16 -8 = -½x2 x = ±4 (-8)÷(-½) = x2 This means that if we were to graph f(x) = -½x2 + 8, as you can see on the next slide, the graph would cross the x-axis at (-4, 0) and (4, 0).

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Graph of f(x) = y = -½x2 + 8

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Solving Equations. -132 = 4x – 5(6x – 10) -132 = 4x – 30x + 50 -132 = -26x + 50 -182 = -26x 7 = x.

Solving Equations. -132 = 4x – 5(6x – 10) -132 = 4x – 30x + 50 -132 = -26x + 50 -182 = -26x 7 = x.

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