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How you find the intercepts using the polynomial? - Answers
I'm assuming you're talking about the equation for a graph here. If you think about the axes on the graph, the y-axis occurs at x = 0, and the x-axis at y = 0. So, to find out where the line crosses the y-axis (or y-intercept), simply set x = 0 in the equation. Example: y = 3x+7 when x = 0, y = 3 x 0 + 7, so y = 7 therefore the y-intercept is 7 To find the x-intercept, simply set y = 0 in the equation. This can be more difficult, especially if it has more than x cubed in it (there are formulae that can be used to solve polynomials, but they can get complicated). Examples: y = 3x+7 when y = 0, 0 = 3x+7, so 3x = -7, and thus x = -7/3 therefore the x-intercept is -7/3 y = x^2+6x+8 when y = 0, x^2+6x+8 = 0, therefore x = -2 or -4 therefore the x-intercepts are -2 and -4 y = x^3+9x^2+30x+24 when y = 0, x^3+9x^2+30x+24 = 0, therefore x = -2, -3 or -4 therefore the x-intercepts are -2, -3 and -4
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How you find the intercepts using the polynomial? - Answers
I'm assuming you're talking about the equation for a graph here. If you think about the axes on the graph, the y-axis occurs at x = 0, and the x-axis at y = 0. So, to find out where the line crosses the y-axis (or y-intercept), simply set x = 0 in the equation. Example: y = 3x+7 when x = 0, y = 3 x 0 + 7, so y = 7 therefore the y-intercept is 7 To find the x-intercept, simply set y = 0 in the equation. This can be more difficult, especially if it has more than x cubed in it (there are formulae that can be used to solve polynomials, but they can get complicated). Examples: y = 3x+7 when y = 0, 0 = 3x+7, so 3x = -7, and thus x = -7/3 therefore the x-intercept is -7/3 y = x^2+6x+8 when y = 0, x^2+6x+8 = 0, therefore x = -2 or -4 therefore the x-intercepts are -2 and -4 y = x^3+9x^2+30x+24 when y = 0, x^3+9x^2+30x+24 = 0, therefore x = -2, -3 or -4 therefore the x-intercepts are -2, -3 and -4
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How you find the intercepts using the polynomial? - Answers
I'm assuming you're talking about the equation for a graph here. If you think about the axes on the graph, the y-axis occurs at x = 0, and the x-axis at y = 0. So, to find out where the line crosses the y-axis (or y-intercept), simply set x = 0 in the equation. Example: y = 3x+7 when x = 0, y = 3 x 0 + 7, so y = 7 therefore the y-intercept is 7 To find the x-intercept, simply set y = 0 in the equation. This can be more difficult, especially if it has more than x cubed in it (there are formulae that can be used to solve polynomials, but they can get complicated). Examples: y = 3x+7 when y = 0, 0 = 3x+7, so 3x = -7, and thus x = -7/3 therefore the x-intercept is -7/3 y = x^2+6x+8 when y = 0, x^2+6x+8 = 0, therefore x = -2 or -4 therefore the x-intercepts are -2 and -4 y = x^3+9x^2+30x+24 when y = 0, x^3+9x^2+30x+24 = 0, therefore x = -2, -3 or -4 therefore the x-intercepts are -2, -3 and -4
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- og:descriptionI'm assuming you're talking about the equation for a graph here. If you think about the axes on the graph, the y-axis occurs at x = 0, and the x-axis at y = 0. So, to find out where the line crosses the y-axis (or y-intercept), simply set x = 0 in the equation. Example: y = 3x+7 when x = 0, y = 3 x 0 + 7, so y = 7 therefore the y-intercept is 7 To find the x-intercept, simply set y = 0 in the equation. This can be more difficult, especially if it has more than x cubed in it (there are formulae that can be used to solve polynomials, but they can get complicated). Examples: y = 3x+7 when y = 0, 0 = 3x+7, so 3x = -7, and thus x = -7/3 therefore the x-intercept is -7/3 y = x^2+6x+8 when y = 0, x^2+6x+8 = 0, therefore x = -2 or -4 therefore the x-intercepts are -2 and -4 y = x^3+9x^2+30x+24 when y = 0, x^3+9x^2+30x+24 = 0, therefore x = -2, -3 or -4 therefore the x-intercepts are -2, -3 and -4
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