find the vertex and y - intercept of the quadratic function - x² + 4x + 7
find the vertex and y - intercept of the quadratic function - x² + 4x + 7
Answer
Explanation:
Step1: Find the vertex
For a quadratic function (y = ax^{2}+bx + c), the (x) - coordinate of the vertex is (x=-\frac{b}{2a}). For the function (y=-x^{2}+4x + 7), where (a=-1), (b = 4), (c = 7). (x=-\frac{4}{2\times(-1)}=\frac{-4}{-2}=2) Substitute (x = 2) into the function: (y=-(2)^{2}+4\times2+7=-4 + 8+7=11) So the vertex is ((2,11))
Step2: Find the (y) - intercept
The (y) - intercept is found by setting (x = 0) in the function. When (x = 0), (y=-(0)^{2}+4\times0+7=7)
Answer:
The vertex is ((2,11)) and the (y) - intercept is (7)