# University of Idaho Quadratic Functions Questions

University of Idaho Quadratic Functions Questions

University of Idaho Quadratic Functions Questions

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### UNFORMATTED ATTACHMENT PREVIEW

The height y (in feet) of a ball thrown by a child is 1 y -X2 + 6x + 5 where x is the horizontal distance in feet from the point at which the ball is thrown. = 16 (a) How high is the ball when it leaves the child’s hand? feet (b) What is the maximum height of the ball? feet (C) How far from the child does the ball strike the ground? Round your answers to the nearest 0.01. feet You are working with a quadratic equation and constructing the following graph. 9H 8 7 7 6 4 3 2 1 -8 -7 -6 -5 -3 -2 – 1 -1 1 2 5 6 7 8 -2 -3 -4 1-5 to Identify the vertex of the parabola: Remember that the vertex is a point! Identify the y-intercept of the parabola: Remember that the y-intercept is a point! Identify the x-intercepts: and Remember that the x-intercepts represent points on the graph! Given the x-intercepts above, write an equation for the parabola in factored form: y = Hint: Think about the zero-product property. Write an equation for the axis of symmetry: Put the equation y = x2 + 12x + 32 into the form y = (x – h)? + k: Answer: y = Graph the function g(x) = 0.5×2 – 4x + 6 on the axes below 15+ 14 13 12 11 10 9 8 7 6 5 4 New 1 1 2 15-14-13-12-11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 -1 3 4 5 6 7 8 9 10 11 12 13 14 15 -2 -3 -4 -5 -6 -7 -8 -9 -10 -11 -12 -13 -14 -15+ Clear All Draw: TAVO Sketch the graph of y = – 1(x – 4)2 + 3. Plot the vertex and another point. +10+ 9 8 7 6 5 4 3 2 1 3 4 5 6 7 8 9 10 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 -1 -2 -3 -4 -5 -6 -7 -8 -9 -10+ Clear All Draw: LAVO Sketch the graph of y = (x + 4)2 – 3. 10+ 9 8 7 6 5 4 3 2 1 -10 -9 -8 -7 -6 -5 -4 -3 -2 4 6 8 9 10 -1 -2 -3 -5 -6 -7 -8 -9 -10+ Clear All Draw: TAN 3. 2 1 -5 -4 -3 -1 1 2 3 4 -2 What is the least possible degree of the polynomial graphed above? Solve the equation: 22 + 6z – 1 = 0. Fully simplify all answers, including non-real solutions. z = Question Help: D Video Consider the parabola given by the equation: f(x) = – 2×2 + 10x – 9 Find the following for this parabola: A) The vertex: B) The vertical intercept is the point C) Find the coordinates of the two X -intercepts of the parabola and write them as a list, separated by commas: It is OK to round your value(s) to to two decimal places. You are working with a quadratic equation and constructing the following graph. 94 8 7 6 5 4 3 2 1 6 7 8 -8 -7 -6 -5 -4 -3 -2 -1 -1 -2 -3 -4 -5 to Identify the vertex of the parabola: Remember that the vertex is a point! Identify the y-intercept of the parabola: Remember that the y-intercept is a point! Identify the x-intercepts: and Remember that the x-intercepts represent points on the graph! Given the x-intercepts above, write an equation for the parabola in factored form: .y = Hint: Think about the zero-product property. Write an equation for the axis of symmetry: Put the equation y = 2 a? – 14x + 48 into the form y = (x – h)? + k: Answer: y = Find the degree of the term – 5×2: Find the degree of the term 2: Find the degree of the term – 6×5: Find the degree of the term 4×4: Find the degree of the polynomial – 5×2 + 2 – 625 + 4×04: Given the function f(a) = (x – 1)(x + 5)(x − 2) its y-intercept is its x-intercepts are The polynomial of degree 5, P(x), has leading coefficient 1, has roots of multiplicity 2 at x = 3 and 0, and a root of multiplicity 1 at x = – 1. X = Find a possible formula for P(a). P(x) = The polynomial of degree 3, P(x), has a root of multiplicity 2 at x = 5 and a root of multiplicity 1 at — 5. The y-intercept is y = – 37.5. X = Find a formula for P(x). P(x) = Write an expression in factored form for the polynomial of least possible degree graphed below. 5+ 4 3 2 -5 -4 -2 2 13 4. 5 -1 -2 -3 -4 -5+ Q y(x) = You are working with a quadratic equation and constructing the following graph. 3 2 1 -8 -7 -6 -5 -4 -3 -2 -1 2 3 4 5 6 7 8 -1 -2 -3 -4 -5+ to Identify the vertex of the parabola: Remember that the vertex is a point! Identify the y-intercept of the parabola: Remember that the y-intercept is a point! Identify the x-intercepts: and Remember that the x-intercepts represent points on the graph! Given the x-intercepts above, write an equation for the parabola in factored form: y = Hint: Think about the zero-product property. Write an equation for the axis of symmetry: Given the function f(x) = (x – 1)(x + 6)(x – 2) its y-intercept is its x-intercepts are Graph the function g(x) = = 0.5×2 – 2x + 3 on the axes below 115 14 13 12 10 ON 8 7 6 5 4 3 2 1 15 -14 -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -4 1 2 3 4 5 6 7 § 9 10 11 12 13 14 15 10 11 12 13 14 1 -2 -3 -4 -5 -6 -7 -8 -9 -10 -11 -12 -13 -14 -15+ Clear All Draw: n Sketch the graph of y = – (x – 2)2 + 5. 10 9 8 8 7 6 6 5 5 4 3 2 1 1 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 -1 1 2 3 4 5 6 7 8 9 10 -2 -3 -4 -5 -6 -7 -8 -9 -10 Clear All Draw: AVO unction Holn: vidad
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University of Idaho Quadratic Functions Questions

20 attachmentsSlide 1 of 20

### UNFORMATTED ATTACHMENT PREVIEW

The height y (in feet) of a ball thrown by a child is 1 y -X2 + 6x + 5 where x is the horizontal distance in feet from the point at which the ball is thrown. = 16 (a) How high is the ball when it leaves the child’s hand? feet (b) What is the maximum height of the ball? feet (C) How far from the child does the ball strike the ground? Round your answers to the nearest 0.01. feet You are working with a quadratic equation and constructing the following graph. 9H 8 7 7 6 4 3 2 1 -8 -7 -6 -5 -3 -2 – 1 -1 1 2 5 6 7 8 -2 -3 -4 1-5 to Identify the vertex of the parabola: Remember that the vertex is a point! Identify the y-intercept of the parabola: Remember that the y-intercept is a point! Identify the x-intercepts: and Remember that the x-intercepts represent points on the graph! Given the x-intercepts above, write an equation for the parabola in factored form: y = Hint: Think about the zero-product property. Write an equation for the axis of symmetry: Put the equation y = x2 + 12x + 32 into the form y = (x – h)? + k: Answer: y = Graph the function g(x) = 0.5×2 – 4x + 6 on the axes below 15+ 14 13 12 11 10 9 8 7 6 5 4 New 1 1 2 15-14-13-12-11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 -1 3 4 5 6 7 8 9 10 11 12 13 14 15 -2 -3 -4 -5 -6 -7 -8 -9 -10 -11 -12 -13 -14 -15+ Clear All Draw: TAVO Sketch the graph of y = – 1(x – 4)2 + 3. Plot the vertex and another point. +10+ 9 8 7 6 5 4 3 2 1 3 4 5 6 7 8 9 10 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 -1 -2 -3 -4 -5 -6 -7 -8 -9 -10+ Clear All Draw: LAVO Sketch the graph of y = (x + 4)2 – 3. 10+ 9 8 7 6 5 4 3 2 1 -10 -9 -8 -7 -6 -5 -4 -3 -2 4 6 8 9 10 -1 -2 -3 -5 -6 -7 -8 -9 -10+ Clear All Draw: TAN 3. 2 1 -5 -4 -3 -1 1 2 3 4 -2 What is the least possible degree of the polynomial graphed above? Solve the equation: 22 + 6z – 1 = 0. Fully simplify all answers, including non-real solutions. z = Question Help: D Video Consider the parabola given by the equation: f(x) = – 2×2 + 10x – 9 Find the following for this parabola: A) The vertex: B) The vertical intercept is the point C) Find the coordinates of the two X -intercepts of the parabola and write them as a list, separated by commas: It is OK to round your value(s) to to two decimal places. You are working with a quadratic equation and constructing the following graph. 94 8 7 6 5 4 3 2 1 6 7 8 -8 -7 -6 -5 -4 -3 -2 -1 -1 -2 -3 -4 -5 to Identify the vertex of the parabola: Remember that the vertex is a point! Identify the y-intercept of the parabola: Remember that the y-intercept is a point! Identify the x-intercepts: and Remember that the x-intercepts represent points on the graph! Given the x-intercepts above, write an equation for the parabola in factored form: .y = Hint: Think about the zero-product property. Write an equation for the axis of symmetry: Put the equation y = 2 a? – 14x + 48 into the form y = (x – h)? + k: Answer: y = Find the degree of the term – 5×2: Find the degree of the term 2: Find the degree of the term – 6×5: Find the degree of the term 4×4: Find the degree of the polynomial – 5×2 + 2 – 625 + 4×04: Given the function f(a) = (x – 1)(x + 5)(x − 2) its y-intercept is its x-intercepts are The polynomial of degree 5, P(x), has leading coefficient 1, has roots of multiplicity 2 at x = 3 and 0, and a root of multiplicity 1 at x = – 1. X = Find a possible formula for P(a). P(x) = The polynomial of degree 3, P(x), has a root of multiplicity 2 at x = 5 and a root of multiplicity 1 at — 5. The y-intercept is y = – 37.5. X = Find a formula for P(x). P(x) = Write an expression in factored form for the polynomial of least possible degree graphed below. 5+ 4 3 2 -5 -4 -2 2 13 4. 5 -1 -2 -3 -4 -5+ Q y(x) = You are working with a quadratic equation and constructing the following graph. 3 2 1 -8 -7 -6 -5 -4 -3 -2 -1 2 3 4 5 6 7 8 -1 -2 -3 -4 -5+ to Identify the vertex of the parabola: Remember that the vertex is a point! Identify the y-intercept of the parabola: Remember that the y-intercept is a point! Identify the x-intercepts: and Remember that the x-intercepts represent points on the graph! Given the x-intercepts above, write an equation for the parabola in factored form: y = Hint: Think about the zero-product property. Write an equation for the axis of symmetry: Given the function f(x) = (x – 1)(x + 6)(x – 2) its y-intercept is its x-intercepts are Graph the function g(x) = = 0.5×2 – 2x + 3 on the axes below 115 14 13 12 10 ON 8 7 6 5 4 3 2 1 15 -14 -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -4 1 2 3 4 5 6 7 § 9 10 11 12 13 14 15 10 11 12 13 14 1 -2 -3 -4 -5 -6 -7 -8 -9 -10 -11 -12 -13 -14 -15+ Clear All Draw: n Sketch the graph of y = – (x – 2)2 + 5. 10 9 8 8 7 6 6 5 5 4 3 2 1 1 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 -1 1 2 3 4 5 6 7 8 9 10 -2 -3 -4 -5 -6 -7 -8 -9 -10 Clear All Draw: AVO unction Holn: vidad
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