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Applying algebraic skills to linear equations . (given the gradient). I can…. Equations of a straight line. …identify gradient and y -intercept from y = mx + c. …identify gradient and a point on the line from y - b = m(x – a).
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Applying algebraic skills to linear equations (given the gradient) I can… Equations of a straight line …identify gradient and y-intercept from y = mx + c …identify gradient and a point on the line from y - b = m(x – a) …use y = mx + c to find the equation of a straight line, given the y-intercept and the gradient …use y – b = m(x – a) to find the equation of a straight line, given one point and the gradient …use functional notation f(x)
…identify gradient and y-intercept from y = mx + c y = mx + c c stands for the y-intercept m stands for the gradient The point where the line cuts the y-axis. Step 1: rewrite the equation in the form y = mx + c Step 2: read off the gradient (m) and the y-intercept (c) Examples The line with the equation y = - 2 + ½x has gradient and passes through ½ (0, -2) y = ½x - 2 The line with the equation y = 5 has gradient and passes through 0 (0, 5) y = 0x + 5
…identify gradient and a point on the line from y - b = m(x – a) y - b = m(x – a) a stands for the x-coordinate of any point on the line b stands for the y-coordinate of any point on the line m stands for the gradient Step 1: rewrite the equation in the form y - b = m(x – a) Step 2: read off the gradient (m) and the point (a,b), taking care with negative values Examples The line with the equation y - 6 = 4(x – 7) has gradient and passes through (7, 6) 4 -2 The line with the equation y + 3 = -2(x – 5) has gradient and passes through (5, -3) y – (-3) = -2(x – 5)
…use y = mx + c to find the equation of a straight line, given the y-intercept and the gradient y = mx + c c stands for the y-intercept m stands for the gradient The point where the line cuts the y-axis. Step 1: Insert the gradient and y-intercept into the equation Step 2: Tidy up the equation Examples The line with gradient 5 through (0, 6) has the equation The line with gradient -2 through (0, 0) has the equation The line with gradient -1 through (0, -2) has the equation y = 5x + 6 y = -2x + 0 y = -2x y = -x + (–2) y = -x – 2 Note: Parallel lines always have the same gradient.
…use y – b = m(x – a) to find the equation of a straight line, given one point and the gradient y - b = m(x – a) b stands for the y-coordinate of any point on the line a stands for the x-coordinate of any point on the line m stands for the gradient Step 1: Insert the gradient and coordinate (a, b) into the equation Step 2: Tidy up the equation Examples The line with gradient 6 through (2, 3) has the equation The line with gradient -7 through (5, 0) has the equation The line with gradient -3 through (-6, -2) has the equation y - 3 = 6(x – 2) y - 0 = -7(x – 5) y = -7(x – 5) y – (-2) = -3(x - (–6)) y + 2 = -3(x + 6)
…use functional notation f(x) The value of y is dependent on x, so y is a function of x, which can be written as f(x) (said f of x) This notation allows for multiple functions to be discussed without confusion by using different letters for each, e.g. f(x), g(x), h(x),… It is also used to show what value requires substitution into the function. Examples For y = 5x + 7, find y if x=1 For f(x) = 5x + 7, find f(1) y = 5 x 1 + 7 y = 12 f(1) = 5 x 1 + 7 f(1) = 12 For f(x) = 5x + 7, find f(-2) For y = 5x + 7, find y if x=-2 f(-2) = 5 x (-2) + 7 f(-2) = -3 y = 5 x (-2) + 7 y = -3
Applying algebraic skills to linear equations (given the gradient) I can… Equations of a straight line …identify gradient and y-intercept from y = mx + c …identify gradient and a point on the line from y - b = m(x – a) …use y = mx + c to find the equation of a straight line, given the y-intercept and the gradient …use y – b = m(x – a) to find the equation of a straight line, given one point and the gradient …use functional notation f(x)
…identify gradient and y-intercept from y = mx + c y = mx + c c stands for the y-intercept m stands for the gradient The point where the line cuts the y-axis. Step 1: rewrite the equation in the form y = mx + c Step 2: read off the gradient (m) and the y-intercept (c) Examples The line with the equation y = - 2 + ½x has gradient and passes through The line with the equation y = 5 has gradient and passes through
…identify gradient and a point on the line from y - b = m(x – a) y - b = m(x – a) a stands for the x-coordinate of any point on the line b stands for the y-coordinate of any point on the line m stands for the gradient Step 1: rewrite the equation in the form y - b = m(x – a) Step 2: read off the gradient (m) and the point (a,b), taking care with negative values Examples The line with the equation y - 6 = 4(x – 7) has gradient and passes through The line with the equation y + 3 = -2(x – 5) has gradient and passes through
…use y = mx + c to find the equation of a straight line, given the y-intercept and the gradient y = mx + c c stands for the y-intercept m stands for the gradient The point where the line cuts the y-axis. Step 1: Insert the gradient and y-intercept into the equation Step 2: Tidy up the equation Examples The line with gradient 5 through (0, 6) has the equation The line with gradient -2 through (0, 0) has the equation The line with gradient -1 through (0, -2) has the equation Note: Parallel lines always have the same gradient.
…use y – b = m(x – a) to find the equation of a straight line, given one point and the gradient y - b = m(x – a) b stands for the y-coordinate of any point on the line a stands for the x-coordinate of any point on the line m stands for the gradient Step 1: Insert the gradient and coordinate (a, b) into the equation Step 2: Tidy up the equation Examples The line with gradient 6 through (2, 3) has the equation The line with gradient -7 through (5, 0) has the equation The line with gradient -3 through (-6, -2) has the equation
…use functional notation f(x) The value of y is dependent on x, so y is a function of x, which can be written as f(x) (said f of x) This notation allows for multiple functions to be discussed without confusion by using different letters for each, e.g. f(x), g(x), h(x),… It is also used to show what value requires substitution into the function. Examples For y = 5x + 7, find y if x=1 For f(x) = 5x + 7, find f(1) For f(x) = 5x + 7, find f(-2) For y = 5x + 7, find y if x=-2