Unit+5+Journals

5.1 quadratic- an equation that uses the exponent ²

vertex- the lowest point of a parabola and also where the line of symmetry crosses

x-intercept- the value of x when y is 0

y-intercept-the value of y when x is 0

increasing- the point(s) when the line of the parabola goes up when you read it from left to right

decreasing-the point(s) when the line of the parabola goes down when you read it from left to right

maximum- the highest point of a parabola

minimum- the lowest point of a parabola

parabola- a U or V shape on a graph with a vertex and a point of symmetry

5.2

Linear functions and quadratic functions have many differences. Linear function equations do not use exponents but quadratic functions do. On a graph, a linear function's points would form a line while a quadratic function's points would form a U or V shaped parabola.

5.3


 * The football created a downward U shape as it flew through the air. It is a U shape because the exponent tells me it is a quadratic equation and it is downward facing because the a-value is negative
 * Vertex = (75, 22)
 * The x-coordinate of the vertex, 75, represents that this is the middle of the total distance the football was across the field
 * The y-coordinate of the vertex, 22, represents that when the football was thrown 75 feet across it was 22 feet in the air

5.4



5.5:
 * The path of the tennis ball goes up and then comes down.
 * h(1) = 27. This means that after 1 seconds, the ball is 27 ft in the air.
 * The y-intercept of h(t) is (0, 3). This means that at 0 seconds after the ball is hit it is 3 ft above the court.
 * The vertex is (1.25, 28) The x-coordinate 1.25 represents how many seconds the ball is in the air and the y-coordinate 28 represents the height of the ball after 1.25 seconds.
 * The x-intercepts of h(t) are (-2.57,0) and (0.07,0). This means that at -2.57 and 0.07 seconds, the tennis ball is still at 0 ft in the air