Please help! Brainliest to the first answer!
1 answer:
You might be interested in
Remember slope is 
. In the graph we can just pick two points on the line to find the rise and run between. Let's say we choose (0,3) and (1,7). Between those points there is 7-3=4 rise and 1-0=1 run. So the slope is 
View the attached photo.
<span>The quadratic function, y = x2, has an x-intercept at the origin = TRUE
</span>
<span>The quadratic function, y = x2 + 3, has an x-intercept at the origin = FALSE
</span>
<span>From x = -2 to x = 0, the average rate of change for both functions is positive = FALSE
When viewing a graph, you start from the left and go to right and from left to right from -2 to 0 the graph is declining.
</span><span>From x = -2 to x = 0, the average rate of change for both functions is negative = TRUE
Both are declining if you read the graph from left to right.
</span><span>For the quadratic function, y = x2, the coordinate (2, 3) is a solution to the equation of the function. = FALSE
</span><span>
y = x^2
If we insert 2 into y = x^2 we get y = 4 but our point (2,3) has y = 3
</span>For the quadratic function, y = x2 + 3, the coordinate (2, 7) is a solution to the equation of the function. = TRUE
If we insert 2 for x, we see that y = x^2 + 3 = y = 2^2 + 3 = y = 7 and our y value in the point (2,7) is 7.
<span>The quadratic function, y = x2, has an x-intercept at the origin = TRUE
</span>
<span>The quadratic function, y = x2 + 3, has an x-intercept at the origin = FALSE
</span>
<span>From x = -2 to x = 0, the average rate of change for both functions is positive = FALSE
When viewing a graph, you start from the left and go to right and from left to right from -2 to 0 the graph is declining.
</span><span>From x = -2 to x = 0, the average rate of change for both functions is negative = TRUE
Both are declining if you read the graph from left to right.
</span><span>For the quadratic function, y = x2, the coordinate (2, 3) is a solution to the equation of the function. = FALSE
</span><span>
y = x^2
If we insert 2 into y = x^2 we get y = 4 but our point (2,3) has y = 3
</span>For the quadratic function, y = x2 + 3, the coordinate (2, 7) is a solution to the equation of the function. = TRUE
If we insert 2 for x, we see that y = x^2 + 3 = y = 2^2 + 3 = y = 7 and our y value in the point (2,7) is 7.